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Full text of "Schaum's Theory and Problems of Theoretical Mechanics"

<A 
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h 

SCHAUM'S OUTLINE OF 1 

THEORY AND PROBLEMS 

OF 

THEORETICAL MECHANICS 

with an introduction to 

Lagrange's Equations 
and Hamiltonian Theory 



(i 



BY 



MURRAY R. SPIEGEL, Ph.D. 

Professor of Mathematics 
Rensselaer Polytechnic Institute 



SCHAUM'S OUTLINE SERIES 

McGRAW-HILL BOOK COMPANY 
New York, St. Louis, San Francisco, Toronto, Sydney 



Copyright © 1967 by McGraw-Hill, Inc. All rights reserved. Printed in the 
United States of America. No part of this publication may be reproduced, 
stored in a retrieval system, or transmitted, in any form or by any means, 
electronic, mechanical, photocopying, recording, or otherwise, without the prior 
written permission of the publisher. 

60232 

8 9 10 11 12 13 14 15 SH SH 7 5 



Preface 



In the 17th century, Sir Isaac Newton formulated his now famous laws of mechanics. 
These remarkably simple laws served to describe and predict the motions of observable 
objects in the universe, including those of the planets of our solar system. 

Early in the 20th century it was discovered that various theoretical conclusions de- 
rived from Newton's laws were not in accord with certain conclusions deduced from theories 
of electromagnetism and atomic phenomena which were equally well founded experimentally. 
These discrepancies led to Einstein's relativistic mechanics which revolutionized the con- 
cepts of space and time, and to quantum mechanics. For objects which move with speeds 
much less than that of light and which have dimensions large compared with those of atoms 
and molecules Newtonian mechanics, also called classical mechanics, is nevertheless quite 
satisfactory. For this reason it has maintained its fundamental importance in science and 
engineering. 

It is the purpose of this book to present an account of Newtonian mechanics and its 
applications. The book is designed for use either as a supplement to all current standard 
textbooks or as a textbook for a formal course in mechanics. It should also prove useful to 
students taking courses in physics, engineering, mathematics, astronomy, celestial me- 
chanics, aerodynamics and in general any field which needs in its formulation the basic 
principles of mechanics. 

Each chapter begins with a clear statement of pertinent definitions, principles and 
theorems together with illustrative and other descriptive material. This is followed by 
graded sets of solved and supplementary problems. The solved problems serve to illustrate 
and amplify the theory, bring into sharp focus those fine points without which the student 
continually feels himself on unsafe ground, and provide the repetition of basic principles 
so vital to effective learning. Numerous proofs of theorems and derivations of basic re- 
sults are included in the solved problems. The large number of supplementary problems 
with answers serve as a complete review of the material of each chapter. 

Topics covered include the dynamics and statics of a particle, systems of particles and 
rigid bodies. Vector methods, which lend themselves so readily to concise notation and to 
geometric and physical interpretations, are introduced early and used throughout the book. 
An account of vectors is provided in the first chapter and may either be studied at the be- 
ginning or referred to as the need arises. Added features are the chapters on Lagrange's 
equations and Hamiltonian theory which provide other equivalent formulations of 
Newtonian mechanics and which are of great practical and theoretical value. 

Considerably more material has been included here than can be covered in most courses. 
This has been done to make the book more flexible, to provide a more useful book of 
reference and to stimulate further interest in the topics. 

I wish to take this opportunity to thank the staff of the Schaum Publishing Company 
for their splendid cooperation. 



M. R. Spiegel 



Rensselaer Polytechnic Institute 
February, 1967 



CONTENTS 



Page 



Chapter 1 VECTORS, VELOCITY AND ACCELERATION 

Mechanics, kinematics, dynamics and statics. Axiomatic foundations of me- 
chanics. Mathematical models. Space, time and matter. Scalars and vectors. 
Vector algebra. Laws of vector algebra. Unit vectors. Rectangular unit vec- 
tors. Components of a vector. Dot or scalar product. Cross or vector product. 
Triple products. Derivatives of vectors. Integrals of vectors. Velocity. Ac- 
celeration. Relative velocity and acceleration. Tangential and normal acceler- 
ation. Circular motion. Notation for time derivatives. Gradient, divergence 
and curl. Line integrals. Independence of the path. Free, sliding and bound 
vectors. 



Chapter 2 NEWTON'S LAWS OF MOTION. WORK, ENERGY 

AND MOMENTUM 33 

Newton's laws. Definitions of force and mass. Units of force and mass. 
Inertial frames of reference. Absolute motion. Work. Power. Kinetic energy. 
Conservative force fields. Potential energy or potential. Conservation of 
energy. Impulse. Torque and angular momentum. Conservation of momentum. 
• Conservation of angular momentum. Non-conservative forces. Statics or equi- 
librium of a particle. Stability of equilibrium. 



Chapter 6 MOTION IN A UNIFORM FIELD. FALLING BODIES 

AND PROJECTILES 

Uniform force fields. Uniformly accelerated motion. Weight and acceleration 
due to gravity. Gravitational system of units. Assumption of a flat earth. 
Freely falling bodies. Projectiles. Potential and potential energy in a uniform 
force field. Motion in a resisting medium. Isolating the system. Constrained 
motion. Friction. Statics in a uniform gravitational field. 



62 



Chapter 4 THE SIMPLE HARMONIC OSCILLATOR AND 

THE SIMPLE PENDULUM 86 

The simple harmonic oscillator. Amplitude, period and frequency of simple 
harmonic motion. Energy of a simple harmonic oscillator. The damped har- 
monic oscillator. Over-damped, critically damped and under-damped motion. 
Forced vibrations. Resonance. The simple pendulum. The two and three 
dimensional harmonic oscillator. 



Chapter 5 CENTRAL FORCES AND PLANETARY MOTION 116 

Central forces. Some important properties of central force fields. Equations 
of motion for a particle in a central field. Important equations deduced from 
the equations of motion. Potential energy of a particle in a central field. Con- 
servation of energy. Determination of the orbit from the central force. Deter- 
mination of the central force from the orbit. Conic sections, ellipse, parabola 
and hyperbola. Some definitions in astronomy. Kepler's laws of planetary 
motion. Newton's universal law of gravitation. Attraction of spheres and 
other objects. Motion in an inverse square field. 



CONTENTS 

Page 

Chapter 6 MOVING COORDINATE SYSTEMS 144 

Non-inertial coordinate systems. Rotating coordinate systems. Derivative 
operators. Velocity in a moving system. Acceleration in a moving system. 
Coriolis and centripetal acceleration. Motion of a particle relative to the earth. 
Coriolis and centripetal force. Moving coordinate systems in general. The 
Foucault pendulum. 



Chapter 7 SYSTEMS OF PARTICLES 165 

Discrete and continuous systems. Density. Rigid and elastic bodies. Degrees 
of freedom. Center of mass. Center of gravity. Momentum of a system of 
particles. Motion of the center of mass. Conservation of momentum. Angular 
momentum of a system of particles. Total external torque acting on a system. 
Relation between angular momentum and total external torque. Conservation 
of angular momentum. Kinetic energy of a system of particles. Work. Po- 
tential energy. Conservation of energy. Motion relative to the center of mass. 
Impulse. Constraints. Holonomic and non-holonomic constraints. Virtual dis- 
placements. Statics of a system of particles. Principle of virtual work. Equi- 
librium in conservative fields. Stability of equilibrium. D'Alembert's principle. 



Chapter 8 APPLICATIONS TO VIBRATING SYSTEMS, 

ROCKETS AND COLLISIONS 194 

Vibrating systems of particles. Problems involving changing mass. Rockets. 
Collisions of particles. Continuous systems of particles. The vibrating string. 
Boundary-value problems. Fourier series. Odd and even functions. Con- 
vergence of Fourier series. 



Chapter 9 PLANE MOTION OF RIGID BODIES 224 

Rigid bodies. Translations and rotations. Euler's theorem. Instantaneous 
axis of rotation. Degrees of freedom. General motion of a rigid body. Chasle's 
theorem. Plane motion of a rigid body. Moment of inertia. Radius of gyra- 
tion. Theorems on moments of inertia. Parallel axis theorem. Perpendicular 
axes theorem. Special moments of inertia. Couples. Kinetic energy and 
angular momentum about a fixed axis. Motion of a rigid body about a fixed 
axis. Principle of angular momentum. Principle of conservation of energy. 
Work and power. Impulse. Conservation of angular momentum. The com- 
pound pendulum. General plane motion of a rigid body. Instantaneous center. 
Space and body centrodes. Statics of a rigid body. Principle of virtual work 
and D'Alembert's principle. Principle of minimum potential energy. Stability. 



Chapter 10 SPACE MOTION OF RIGID BODIES 253 

General motion of rigid bodies in space. Degrees of freedom. Pure rotation 
of rigid bodies. Velocity and angular velocity of a rigid body with one point 
fixed Angular momentum. Moments of inertia. Products of inertia. Moment 
of inertia matrix or tensor. Kinetic energy of rotation. Principal axes of 
inertia. Angular momentum and kinetic energy about the principal axes. The 
ellipsoid of inertia. Euler's equations of motion. Force free motion. The in- 
variable line and plane. Poinsot's construction. Polhode. Herpolhode. Space 
and body cones. Symmetric rigid bodies. Rotation of the earth. The Euler 
angles. Angular velocity and kinetic energy in terms of Euler angles. Motion 
of a spinning top. Gyroscopes. 



CONTENTS 



Page 

Chapter // LAGRANGE'S EQUATIONS 282 

General methods of mechanics. Generalized coordinates. Notation. Trans- 
formation equations. Classification of mechanical systems. Scleronomic and 
rheonomic systems. Holonomic and non-holonomic systems. Conservative and 
non-conservative systems. Kinetic energy. Generalized velocities. Gener- 
alized forces. Lagrange's equations. Generalized momenta. Lagrange's equa- 
tions for non-holonomic systems. Lagrange's equations with impulsive forces. 



Chapter 12 HAMILTONIAN THEORY 311 

Hamiltonian methods. The Hamiltonian. Hamilton's equations. The Hamil- 
tonian for conservative systems. Ignorable or cyclic coordinates. Phase space. 
Liouville's theorem. The calculus of variations. Hamilton's principle. Can- 
onical or contact transformations. Condition that a transformation be canoni- 
cal. Generating functions. The Hamilton-Jacobi equation. Solution of the 
Hamilton-Jacobi equation. Case where Hamiltonian is independent of time. 
Phase integrals. Action and angle variables. 



APPENDIX A 



UNITS AND DIMENSIONS 339 



APPENDIX B 



ASTRONOMICAL DATA 342 



appendix c SOLUTIONS OF SPECIAL DIFFERENTIAL EQUATIONS .... 344 



APPENDIX D INDEX OF SPECIAL SYMBOLS AND NOTATIONS 356 



INDEX 361 



and ACCELERATION 



MECHANICS, KINEMATICS, DYNAMICS AND STATICS 

Mechanics is a branch of physics concerned with motion or change in position of 
physical objects. It is sometimes further subdivided into: 

1. Kinematics, which is concerned with the geometry of the motion, 

2. Dynamics, which is concerned with the physical causes of the motion, 

3. Statics, which is concerned with conditions under which no motion is apparent. 

AXIOMATIC FOUNDATIONS OF MECHANICS 

An axiomatic development of mechanics, as for any science, should contain the following 
basic ingredients: 

1. Undefined terms or concepts. This is clearly necessary since ultimately any 
definition must be based on something which remains undefined. 

2. Unproved assertions. These are fundamental statements, usually in mathematical 
form, which it is hoped will lead to valid descriptions of phenomena under study. 
In general these statements, called axioms or postulates, are based on experimental 
observations or abstracted from them. In such case they are often called laws. 

3. Defined terms or concepts. These definitions are given by using the undefined 
terms or concepts. 

4. Proved assertions. These are often called theorems and are proved from the 
definitions and axioms. 

An example of the "axiomatic way of thinking" is provided by Euclidean geometry in 
which point and line are undefined concepts. 

MATHEMATICAL MODELS 

A mathematical description of physical phenomena is often simplified by replacing 
actual physical objects by suitable mathematical models. For example in describing the 
rotation of the earth about the sun we can for many practical purposes treat the earth 
and sun as points. 

SPACE, TIME AND MATTER 

From everyday experience, we all have some idea as to the meaning of each of the 
following terms or concepts. However, we would certainly find it difficult to formulate 
completely satisfactory definitions. We take them as undefined concepts. 



2 VECTORS, VELOCITY AND ACCELERATION [CHAP. 1 

1. Space. This is closely related to the concepts of point, position, direction and 
displacement. Measurement in space involves the concepts of length or distance, 
with which we assume familiarity. Units of length are feet, meters, miles, etc. 
In this book we assume that space is Euclidean, i.e. the space of Euclid's geometry. 

2. Time. This concept is derived from our experience of having one event taking 
place after, before or simultaneous with another event. Measurement of time is 
achieved, for example, by use of clocks. Units of time are seconds, hours, years, etc. 

3. Matter. Physical objects are composed of "small bits of matter" such as atoms 
and molecules. From this we arrive at the concept of a material object called a 
particle which can be considered as occupying a point in space and perhaps moving 
as time goes by. A measure of the "quantity of matter" associated with a particle 
is called its mass. Units of mass are grams, kilograms, etc. Unless otherwise 
stated we shall assume that the mass of a particle does not change with time. 

Length, mass and time are often called dimensions from which other physical quantities 
are constructed. For a discussion of units and dimensions see Appendix A, Page 339. 



SCALARS AND VECTORS 

Various quantities of physics, such as length, mass and time, require for their specifica- 
tion a single real number (apart from units of measurement which are decided upon in 
advance). Such quantities are called scalar s and the real number is called the magnitude 
of the quantity. A scalar is represented analytically by a letter such as t, m, etc. 

Other quantities of physics, such as displacement, require for their specification a 
direction as well as magnitude. Such quantities are called vectors. A vector is repre- 
sented analytically by a bold faced letter such as A in Fig. 1-1. Geometrically it is 
represented by an arrow PQ where P is called the initial point and Q is called the terminal 
point. The magnitude or length of the vector is then denoted by |A| or A. 





Fig. 1-1 Fig. 1-2 Fig. 1-3 

VECTOR ALGEBRA 

The operations of addition, subtraction and multiplication familiar in the algebra of 
real numbers are with suitable definition capable of extension to an algebra of vectors. 
The following definitions are fundamental. 

1. Two vectors A and B are equal if they have the same magnitude and direction 
regardless of their initial points. Thus A = B in Fig. 1-2 above. 

2. A vector having direction opposite to that of vector A but with the same length is 
denoted by -A as in Fig. 1-3 above. 

3. The sum or resultant of vectors A and B of Fig. l-4(a) below is a vector C formed 
by placing the initial point of B on the terminal point of A and joining the initial 
point of A to the terminal point of B [see Fig. l-4(b) below]. We write C = A + B. 
This definition is equivalent to the parallelogram law for vector addition as indicated 
in Fig. l-4(c) below. 



CHAP. 11 



VECTORS, VELOCITY AND ACCELERATION 






Fig. 1-4 

Extensions to sums of more than two vectors are immediate. For example, 
Fig. 1-5 below shows how to obtain the sum or resultant E of the vectors A, B, C 
and D. 



5. 




rrfB+c+D 



Fig. 1-5 



The difference of vectors A and B, represented by A — B, is that vector C which 
when added to B gives A. Equivalently, A-B may be denned as A + (-B). If 
A = B, then A — B is defined as the null or zero vector represented by 0. This has 
a magnitude of zero but its direction is not defined. 

The product of a vector A by a scalar p is a vector pA or Ap with magnitude 
\p\ times the magnitude of A and direction the same as or opposite to that of A 
according as p is positive or negative. If p = 0, pA = 0, the null vector. 



LAWS OF VECTOR ALGEBRA 

If A, B and C are vectors, and p and q are scalars, then 



1. A + B = B + A 

2. A + (B + C) = (A + B) + C 

3. p(qA) = (pq)A = q(pA) 

4. (p + q)A = pA + qA 

5. p{A + B) = pA + pB 



Commutative Law for Addition 
Associative Law for Addition 
Associative Law for Multiplication 
Distributive Law 
Distributive Law 



Note that in these laws only multiplication of a vector by one or more scalars is 
defined. On pages 4 and 5 we define products of vectors. 

UNIT VECTORS 

Vectors having unit length are called unit vectors. If A is a vector with length A > 0, 
then A/A = a is a unit vector having the same direction as A and A = A a. 



RECTANGULAR UNIT VECTORS 

The rectangular unit vectors i, j and k are mutually perpendicular unit vectors having 
directions of the positive x, y and z axes respectively of a rectangular coordinate system 



VECTORS, VELOCITY AND ACCELERATION 



[CHAP. 1 



[see Fig. 1-6]. We use right-handed rectangular coordinate 
systems unless otherwise specified. Such systems derive 
their name from the fact that a right threaded screw ro- 
tated through 90° from Ox to Oy will advance in the posi- 
tive z direction. In general three vectors A, B and C which 
have coincident initial points and are not coplanar are said 
to form a right-handed system or dextral system if a right 
threaded screw rotated through an angle less than 180° from 
A to B will advance in the direction C [see Fig. 1-7 below]. 



\/o. 



w v 



Fig. 1-6 




(4i,A2,As) 




Fig. 1-7 



Fig. 1-8 



COMPONENTS OF A VECTOR 

Any vector A in 3 dimensions can be represented with initial point at the origin O of 
a rectangular coordinate system [see Fig. 1-8 above]. Let (Ai,A 2 ,A 3 ) be the rectangular 
coordinates of the terminal point of vector A with initial point at 0. The vectors Aii, 
A 2 j and A 3 k are called the rectangular component vectors, or simply component vectors, 
of A in the x, y and z directions respectively. A lf A 2 and A 3 are called the rectangular 
components, or simply components, of A in the x, y and z directions respectively. 

The sum or resultant of Aii, A 2 j and Ask is the vector A, so that we can write 



A = Aii + A 2 j + A 3 k 



(1) 



The magnitude of A is 



A = |A| = v^f+AfTAf 



In particular, the position vector or radius vector r from O to the point {x,y,z) is 

written . , . , . 

r = xi + yj + zk 



(2) 

is 

(3) 



and has magnitude r = |r| = y/x 2 + y 2 + z 2 . 



DOT OR SCALAR PRODUCT 

The dot or scalar product of two vectors A and B, denoted by A-B (read A dot B) 
is defined as the product of the magnitudes of A and B and the cosine of the angle 
between them. In symbols, 

A«B = AS cos 0, 0^0^77 (4) 



Note that A*B is a scalar and not a vector. 



CHAP. 11 



VECTORS, VELOCITY AND ACCELERATION 



The following laws are valid: 

1. A • B = B • A Commutative Law for Dot Products 

2. A«-(B + C) = A»B + A«C Distributive Law 

3. p{A • B) = (pA) • B = A • (pB) = (A • B)p, where p is a scalar. 

4. i*i = j*j = k*k = l, i*j = j*k = k*i = 

5. If A = Aii + Aaj + Aak and B = Bd + B 2 j + Bak, then 

A • B = AiBi + A2B2 + A3B3 
A- A = A 2 = A\ + A\ + A\ 
B-B = B 2 = B\ + B\ + B% 

6. If A-B = and A and B are not null vectors, then A and B are perpendicular. 



CROSS OR VECTOR PRODUCT 

The cross or vector product of A and B is a vector C = A x B (read A cross B). The 
magnitude of A x B is denned as the product of the magnitudes of A and B and the sine 
of the angle between them. The direction of the vector C = A x B is perpendicular to the 
plane of A and B and such that A, B and C form a right-handed system. In symbols, 

A X B = AB sin 6 u, ^ 6 ^ tt (5) 

where u is a unit vector indicating the direction of A x B. If A = B or if A is parallel 
to B, then sin 6 = and we define A x B = 0. 

The following laws are valid: 



1. 
2. 
3. 
4. 
5. 



A x B = — B x A (Commutative Law for Cross Products Fails) 

Ax(B + C) = AxB + AxC Distributive Law 

p{A x B) = {pA) x B = A x (pB) - (A x B)p, where p is a scalar. 

ixi = jxj = kxk = 0, ixj = k, j x k = i, kxi = j 

If A = Aii + Aaj + Aak and B = Bd + B 2 j + 



AxB = 



1 

At 
Bx 



1 

A 2 

B 2 



then 
k 
A 3 
B 3 



6. 

7. 



|A x B| = the area of a parallelogram with sides A and B. 

If A x B = and A and B are not null vectors, then A and B are parallel. 



TRIPLE PRODUCTS 

The scalar triple product is defined as 





At 


A 2 


As 


• (B x C) = 


Bx 


B 2 


B z 




Ct 


C 2 


Cz 



(') 



where A = Aii + A 2 j + A3k, B = B\\ + B 2 j + Bak, C = Cd + C 2 j + Cak. It represents the 
volume of a parallelepiped having A, B, C as edges, or the negative of this volume according 
as A, B, C do or do not form a right handed system. We have A • (B x C) = B • (C x A) = 

C-(AXB). 

The vector triple product is defined as 

AX(BXC) = (A-C)B-(A-B)C {7) 

Since (A x B) x C = (A • C)B - (B • C) A, it is clear that A x (B x C) ¥• (A x B) x C. 



6 VECTORS, VELOCITY AND ACCELERATION [CHAP. 1 

DERIVATIVES OF VECTORS 

If to each value assumed by a scalar variable u there corresponds a vector A(u), or 
briefly A, then A(u) is called a (vector) function of u. The derivative of A(u) is defined as 

dU Au~0 AW V ' 

provided this limit exists. If A(u) = Ai(u)i + Ai{u)i + A 3 (w)k, then 

dA _ dAi . dAi. c?A 3 . ,q* 

du ~~ du du du 

Similarly we can define higher derivatives. For example the second derivative of A(u) 

if it exists is given by 

&A _ ^Ai. &A2. d?As , /m 

du 2 ~ du* 1 + du 2 3 + « K I 1 "* 

Example. If A = (2m 2 — 3tt)i + 5 cos u j — 3 sin u k, then 

^ = (4m - 3)i - 5 sin m j — 3 cos u k, -j-g- = 4i - 5 cos u j + 3 sin u k 

The usual rules of differentiation familiar in the calculus can be extended to vectors, 
although order of factors in products may be important. For example if ${u) is a scalar 
function while A{u) and B(u) are vector functions, then 

t(*-*) = *•£ + £•» <"> 

A(AXB) = Axf + fxB (IS) 

INTEGRALS OF VECTORS 

Let A{u) = Ai(u)i + A 2 (u)j + A s (u)k be a vector function of u. We define the indefinite 
integral of A(u) as 

( A(u)du = i f A x {u)du + j J A 2 (w)^ + kj A 3 (u)du (U) 

If there exists a vector function B(u) such that A(w) = ^-{B(w)}, then 

J*A(M)dM = r^{B(tt)}<fa = B(m) + c (*5) 

where c is an arbitrary constant vector independent of u. The definite integral between 
limits u = a and u = /? is in such case, as in elementary calculus, given by 

C A(u)du = f'-fo{B{u))du = B(u) + c ' = B(/3) - B(«) (iff) 

The definite integral can also be defined as a limit of a sum analogous to that of elementary 
calculus. 

VELOCITY 

Suppose that a particle moves along a path or curve C [Fig. 1-9 below]. Let the position 
vector of point P at time t be r = r(t) while the position vector of point Q at time t + At is 



CHAP. 11 



VECTORS, VELOCITY AND ACCELERATION 



r + Ar = r(t + At). Then the velocity (also called the instantaneous velocity) of the particle 
at P is given by 



dr .. Ar 

v = -— = hm — - 
at At-*o At 

At-*o AC 

and is a vector tangent to C at P. 



(17) 



If r = r(t) = x(t)i + y(t)j + z{t)k = xi + yj + zk, 
we can write 

dt dx. dy . dz . . 

v = dt = dt 1 + ^ 3 + Tt k ( 18) 



The magnitude of the velocity is called the speed and is given by 




Fig. 1-9 



dr 
dt 



- V(t y+ (f r ♦ (t ) 2 = 



V = |v| = 
where s is the arc length along C measured from some initial point to P. 



dt 



(19) 



ACCELERATION 

If v = dr/dt is the velocity of the particle, we define the acceleration (also called the 
instantaneous acceleration) of the particle at P as 



a = dv = Hm v(t + At)-v(t) 

dt At-»0 At 



In terms of r = xi + yj + zk the acceleration is 

_ d^r _ d?x. d?y. 
* ~ dt* ~ dt 21 + dt 2 ~ 

and its magnitude is 



i + 5^J + ^5k 



dV 

dt 2 ' 



a = 



= V(S)' + (SWS 



(21) 
(22) 



RELATIVE VELOCITY AND ACCELERATION 

If two particles Pi and Pz are moving with respective velocities vi and V2 and accelera- 
tions ai and a2, the vectors 

vpg/Pj = v 2 - vi and ap 2 /Pj = a2 - ai (23 

are respectively called the relative velocity and relative acceleration of P2 with respect to Pi 



TANGENTIAL AND 
NORMAL ACCELERATION 

Suppose that particle P with position vec- 
tor r = r(t) moves along curve C [Fig. 1-10]. 
We can consider a rectangular coordinate 
system moving with the particle and defined 
by the unit tangent vector T, the unit princi- 
pal normal N and the unit binormal B to 
curve C where 




Fig. 1-10 



8 



VECTORS, VELOCITY AND ACCELERATION 



[CHAP. 1 



T = 



B = TxN 



(U) 



* N = R M 
ds' * K ds' 

s being the arc length from some initial point to P and R the radius of curvature of C at P. 
The reciprocal of the radius of curvature is called the curvature and is given by k= 1/R. 

We can show [see Problem 1.35, page 20] that the acceleration along C is given by 



dt L + R* 1 



(25) 



The first and second terms on the right are called the tangential acceleration and normal 
or centripetal acceleration respectively. 



CIRCULAR MOTION 

Suppose particle P moves on a circle C of radius 
R. If s is the arc length measured along C from 
A to P and 9 is the corresponding angle subtended 
at the center O, then s = Rd. Thus the magnitudes 
of the tangential velocity and acceleration are given 
respectively by 

ft = fi » (« e > 



v 



at 



and 



dv _ d?s _ p d 2 
dt dt 2 K dt 2 




= Ra 



(27) 



Fig. 1-11 



We call © = dd/dt and a - d 2 9/dt 2 the angular speed and angular acceleration respectively. 
The normal acceleration as seen from (25) is given by v 2 /R = <a 2 R. 

NOTATION FOR TIME DERIVATIVES 

We shall sometimes find it convenient to use dots placed over a symbol to denote 
derivatives with respect to time t, one dot for a first derivative, two dots for a second 
derivative, etc. Thus for example r = dr/dt, r = d 2 r/dt 2 , v = dvldt, etc. 



GRADIENT, DIVERGENCE AND CURL 

If to each point (x, y, z) of a rectangular coordinate system there corresponds a vector A, 
we say that A = A(x, y, z) is a vector function of x, y, z. We also call A(x, y, z) a vector 
field. Similarly we call the (scalar) function <f>(x, y, z) a scalar field. 

It is convenient to consider a vector differential operator called del given by 



Using this we define the following important quantities. 

= ( 1 a^ + J ^ + k ^ 



1. Gradient V4> = (i^ + J^7 + k ^j<^ - * dx 

This is a vector called the gradient of <£ and is also written grad </>. 



+ 3 dy + K dz 



(28) 



(29) 



2. Divergence 



V'A = 



l Jx + z dy dz 

Mi + Mi + M§ 
a« a?/ dz 



(Ad + A 2 j + AOt) 



This is a scalar called the divergence of A and is also written div A. 



(SO) 



CHAP. 1] 



VECTORS, VELOCITY AND ACCELERATION 



3. Curl 



V x A = 



i 5S + ^ + k £) x < A ' i + A2i + A * ) 



i 


j 


k 


d 


d 


d 


dX 


dy 


dz 


At 


A 2 


As 



(31) 



fdAs 
\dy 



BA 
dz 



„ l + ( g 



dA s \. , fdA 2 



dX 



j + [iF-f ]k 



This is a vector called the curl of A and is also written curl A. 
Two important identities are 

divcurlA = V(V'XA) = 
curl grad <£ = V x (V4>) = 



(32) 



LINE INTEGRALS 

Let r(t) = x(t)i + y(t)j + z(t)k, where r(t) is the position vector of (x,y,z), define a 
curve C joining points Pi and Pi corresponding to t = U and t = t 2 respectively. Let 
A = A(x, y, z) = Ad + A 2 j + Ask be a vector function of position (vector field). The 
integral of the tangential component of A along C from Pi to P%, written as 



JA'dr = ( A«dr = I Aidx + A 2 dy + A 3 dz 



(84) 



is an example of a line integral. 

If C is a closed curve (which we shall suppose is a simple closed curve, i.e. a curve 
which does not intersect itself anywhere) then the integral is often denoted by 



£ A-dr = <k Aidx + A 2 dy + A s dz 



(35) 



In general, a line integral has a value which depends on the path. For methods of 
evaluation see Problems 1.39 and 1.40. 



INDEPENDENCE OF THE PATH 

The line integral (34) will be independent of the path joining Pi and P 2 if and only if 
A = v<£, or equivalently V x A = 0. In such case its value is given by 



J»p 2 x»p 2 

A-dr = I d<f> = <j>(P 2 ) - 4>(Pi) = <f>(x 2 ,y 2 ,z 2 ) - <f>(x u yi,Zx) 
Pt J Pi 



(36) 



assuming that the coordinates of Pi and P 2 are (x v y v z t ) and (x 2 , y 2 , z 2 ) respectively while 
<f>(x,y,z) has continuous partial derivatives. The integral (35) in this case is zero. 



FREE, SLIDING AND BOUND VECTORS 

Up to now we have dealt with vectors which are specified by magnitude and direction 
only. Such vectors are called free vectors. Any two free vectors are equal as long as 
they have the same magnitude and direction [see Fig. l-12(a) below]. 



10 



VECTORS, VELOCITY AND ACCELERATION 



[CHAP. 1 



(a) Equal free vectors 



(b) Equal sliding vectors 
Fig. 1-12 



(c) Bound vector 



Sometimes in practice the particular line of action of a vector is important. In such 
case two vectors are equal if and only if they have the same magnitude, direction and line 
of action. Such vectors are often called sliding vectors [see Fig. 1-12(6)]. 

Sometimes it is important to specify the point of action of a vector. Such a vector 
[see Fig. l-12(c)] is called a bound vector. In this case two vectors will be equal if and 
only if they are identical. 

Most cases with which we shall deal involve free vectors. Cases where sliding vectors 
or bound vectors need to be employed will in general be clear from the context. 



Solved Problems 

VECTOR ALGEBRA 

1.1. Show that addition of vectors is commutative, i.e. A + B = B + A. See Fig. 1-13 
below. 

OP + PQ . = OQ or A + B = C 

and OR + RQ = OQ or B + A = C 

Then A + B = B + A. 





1.2. Show that the addition of vectors is associative, i.e. A + (B + C) = (A 4- B) + C. See 
Fig. 1-14 above. 

OP + PQ = OQ = (A + B) and PQ + QR = PR = (B + C) 

Since OP + PR = OR = D, i.e. A + (B + C) = D 

OQ + QR = OR = D, i.e. ( A + B) + C = D 

we have A + (B + C) = ( A + B) + C. 

Extensions of the results of Problems 1.1 and 1.2 show that the order of addition of any 
number of vectors is immaterial. 



CHAP. 1] 



VECTORS, VELOCITY AND ACCELERATION 



11 



1.3. Given vectors A, B and C [Fig. l-15(a)] construct (a) A - B + 2C, (b) 3C - £(2A - B). 




(a) 




-B 




Pig. 1-15 

1.4. Prove that the magnitude A of the vector A = 
Ad + Aaj + Ask is A = y/A\ + A\ + A\. See 
Fig. 1-16. 

By the Pythagorean theorem, 

(OP)* = (OQ)* + (QP) 2 
where OP denotes the magnitude of vector OP, etc. 
Similarly, (OQ)* = (OR)* + (RQ)*. 

Then (OP) 2 = (OR)* + (RQ)* + (QP)* or 
A* = A\ + A\ + A\, i.e. A = VA* + A* + A* 



-|(2A-B) 




1.5. Determine the vector having the initial point 
P(xi,yi,zi) and the terminal point Q(* 2 , 3/2, z 2 ) 
and find its magnitude. See Fig. 1-17. 

The position vector of P is r t = x x \ + 2/ x j + Zjk. 

The position vector of Q is r 2 = x 2 i + 2/ 2 J + «2 k - 

r x + PQ = r 2 or 

PQ = 1*2 — r i = (^2* + Vzi + *2 k ) — (M + #ii + *i k ) 
= («2 - *i)i + (Va - 2/i)J + («2 - «i)k 

Magnitude of PQ = PQ 



= V(*2 - *l) 2 + (^2 ~ Vl) 2 + (*2 ~ *l) 2 

Note that this is the distance between points P and Q. 




Fig. 1-17 



12 



VECTORS, VELOCITY AND ACCELERATION 



[CHAP. 1 



1.6. Find (a) graphically and (b) analytically the sum or resultant of the following 
displacements: 

A, 10 ft northwest; B, 20 ft 30° north of east; C, 35 ft due south. See Fig. 1-18. 

Graphically. 

At the terminal point of A place the initial point 
of B. At the terminal point of B place the initial 
point of C. 

The resultant D is formed by joining the initial 
point of A to the terminal point of C, i.e. D = 

A + B + C. 

The resultant is measured to have magnitude of 
4.1 units = 20.5 ft and direction 60° south of east. 

Analytically. 

From Fig. 1-18, if i and j are unit vectors in the 
E and N directions, we have 

A = - 10 cos 45° i + 10 sin 45° j 

B = 20 cos 30° i + 20 sin 30° j 

C = — 35j 

Then the resultant is 

D = A + B + C = (-10 cos 45° + 20cos30°)i + (10 sin 45° + 20 sin 30° - 35)j 
= (-5V2 + 10VS )i + (5 V^ + 10 - 35)j = 10.25i - 17.93J 




Unit = 5 ft 



Thus the magnitude of D is V(10.25) 2 + (17.93) 2 = 20.65 ft and the direction is 

tan-i 17.93/10.25 = tan" 1 1.749 = 60°45' south of east 

Note that although the graphical and analytical results agree fairly well, the analytical result is 
of course more accurate. 



THE DOT OR SCALAR PRODUCT 

1.7. Prove that the projection of A on B is equal to A • b, 
where b is a unit vector in the direction of B. 

Through the initial and terminal points of A pass 
planes perpendicular to B at G and H respectively as in the 
adjacent Fig. 1-19; then 

Projection of A on B = GH = EF = A cos 6 = A • b 




G H B 

Fig. 1-19 



1.8. Prove A«(B + C) = AB + AC. 

Let a be a unit vector in the direction of A; then [see 
Fig. 1-20] 

Projection of (B + C) on A = projection of B on A 

+ projection of C on A 

(B + C)»a = B*a + C*a 

Multiplying by A, 

(B+C)'Aa = B»Aa + OAa 

and (B + C) • A = B • A + C • A 

Then by the commutative law for dot products, 

A • (B + C) = A • B + A • C 
and the distributive law is valid. 




Fig. 1-20 



CHAP. 1] 



VECTORS, VELOCITY AND ACCELERATION 



13 



1.9. Evaluate each of the following. 

(a) i«i = |i||i|cosO° = (1)(1)(1) = 1 
(6) i»k = |i||k|cos90° = (1)(1)(0) = 
(e) k-j = |k||j|cos90° = (1)(1)(0) = 

(d) j • (2i - 8j + k) = 2j • i - 3j • j + j • k = 0-3 + = -3 

(e) (2i - j) • (3i + k) = 2i • (3i + k) - j • (3i + k) = 6i • i + 2i • k - 3j • i - j • k 

= 6 + 0-0-0 = 6 

1.10. If A = Aii + Aaj+Agk and B = Bd + B 2 2 + B^k, prove that A-B = Ai#i + A 2 JB 2 + 

AsBa. 

A-B = (Aii + Aa + AM-iBj + Bzi + Bak) 

= A x \ • (BJ + Baj + B 3 k) + A 2 j • (£ x i + B 2 j + B 3 k) + A 3 k • (BJ + B 2 j + B 3 k) 

= A x B x i • i + A^i • j + A X B Z \ • k + A 2 Btf • i + A 2 £ 2 j • j + A 2 £ 3 j • k 

+ A^k • i + A 3 £ 2 k • j + A 3 £ 3 k • k 

= A X B X + A 2 B % + A 3 fi 3 

since i • i = j • j = k • k = 1 and all other dot products are zero. 



1.11. If A = Aii + A 2 j + A 3 k, show that A = \/A-A = y/A\ + Al + A\. 

A«A = (A)(A) cos 0° = A 2 . Then A = VA • A. 

Also, A»A = (Aji + A 2 j + A 3 k) • (A 1 i + A 2 j + A 3 k) 

= (AJiAJ + (A 2 )(A 2 ) + (A S )(A S ) = A\ + A\ + Al 
by Problem 1.10, taking B = A. 

Then A = y/A* A = -^A\ + A 2 + A\ is the magnitude of A. Sometimes A • A is written A 2 . 



1.12. Find the acute angle between the diagonals of a 
quadrilateral having vertices at (0, 0, 0), (3, 2, 0), 

(4,6,0), (1,3,0) [Fig. 1-21]. 

We have OA = 3i + 2j, OB = 4i + 6j, OC = i + 3j 
from which 

CA = OA-OC = 2i — j 

Then OB • CA = [OB| |CA| cos $ 
i.e. 

(4i + 6j) • (2i - j) = V(4) 2 + (6)2 V(2) 2 + (-1) 2 cos e 

from which cos 6 = 2/(v1S2 Vs ) = -1240 and = 82°53'. 



5(4,6,0) 




(0,0,0) 



Fig. 1-21 



THE CROSS OR VECTOR PRODUCT 
1.13. Prove AxB = -BxA. 




Fig. 1-22 




BXA = D 



(&) 



A X B = C has magnitude AB sin $ and direction such that A, B and C form a right-handed 
system [Fig. l-22(a) above]. 



14 



VECTORS, VELOCITY AND ACCELERATION 



[CHAP. 1 



B X A = D has magnitude BA sin e and direction such that B, A and D form a right-handed 
system [Pig. 1-22(6) above]. 

Then D has the same magnitude as C but is opposite in direction, i.e. C = — D or A X B = -B X A. 

The commutative law for cross products is not valid. 



1.14. Prove that 



Ax(B + C) = AXB + AXC 

for the case where A is perpendicular 
to B and also to C. 

Since A is perpendicular to B, A X B is a 
vector perpendicular to the plane of A and B 
and having magnitude AB sin 90° = AB or 
magnitude of AB. This is equivalent to mul- 
tiplying vector B by A and rotating the 
resultant vector through 90° to the position 
shown in Fig. 1-23. 

Similarly, A X C is the vector obtained by 
multiplying C by A and rotating the resultant 
vector through 90° to the position shown. 

In like manner, A X (B + C) is the vector 
obtained by multiplying B + C by A and rotat- 
ing the resultant vector through 90° to the 
position shown. 

Since A X (B + C) is the diagonal of the 
parallelogram with A X B and A X C as sides, 
we have AX(B-f-C) = AXB + AXC. 







Fig. 1-23 




1.15. Prove that A x (B + C) = AxB + AxC 
in the general case where A, B and C are 
non-coplanar. See Fig. 1-24. 

Resolve B into two component vectors, one 
perpendicular to A and the other parallel to A, 
and denote them by B^ and B ( | respectively. 
Then B = B ± + B„. 

If 6 is the angle between A and B, then 
B j_ = B sin e. Thus the magnitude of A X B j_ is 
AB sin 9, the same as the magnitude of A X B. 
Also, the direction of A X B ± is the same as the 
direction of A X B. Hence A X B ± =AXB. 

Similarly if C is resolved into two component 
vectors C ( | and C^, parallel and perpendicular 
respectively to A, then A X C ± = A X C. Fig. 1-24 

Also, since B + C = B ± + B n + C ± + C M = (Bj_ + C ± ) + (B, ( + C M ) it follows that 

AX(B ± + C 1 ) = AX(B + C) 

Now Bjl and Cj_ are vectors perpendicular to A and so by Problem 1.14, 

AX(B 1 + C 1 ) = AXB i +AXC 1 

Then A X (B + C) = AXB + AXC 

and the distributive law holds. Multiplying by —1, using Problem 1.13, this becomes (B + C) X A = 
B X A + C X A. Note that the order of factors in cross products is important. The usual laws of 
algebra apply only if proper order is maintained. 

i 3 k 

1.16. If A = Ad + A 2 j + Ask and B = Bii + B 2 j + Bak, prove that A x B 



Ai 



A 2 
B 2 



As 
Bs 



CHAP. 1] 



VECTORS, VELOCITY AND ACCELERATION 



15 



AXB = (Aii + Atf + AMxiBii + Btf + BJi.) 

= Aii X (S x i + Ba + B s k) + Aa X (BJ + Brf + £ 3 k) + A 3 k X (BJ + # 2 j + B 3 k) 
= A^i X i + A x B 2 i X j + A X B Z \ X k + AgBJ X i + A 2 5 2 j X J + ^2#3J X k 
+ A 3 Bik X i + A 3 B 2 k X j + A 3 B 3 k X k 



= (A 2 B 3 - A 3 B 2 )i + (A s B t - A X B Z )\ + (A t B 2 - A^k = 

1.17. If A = 3i-j + 2k and B = 2i + 3j-k, findAxB. 

j k 



i 

At 
B, 



j 


k 


A 2 


^3 


B 2 


*3 



AXB = 



-1 2 
3 -1 





-1 


2 




3 


2 




3 


-1 


= 1 






— J 






+ k 








3 


-1 




2 


-1 




2 


3 



= -5i + 7j + Ilk 



1.18. Prove that the area of a parallelogram with 
sides A and B is |AxB|. 

Area of parallelogram = h |B| 

= | A| sin 6 |B| 

= |AXB| 

Note that the area of the triangle with sides A and 
B is £ |A X B| . 




B 

Fig. 1-25 



= £|19i-4j + 7k| 



1.19. Find the area of the triangle with vertices at P(2,3,5), Q(4,2,-l), 72(3,6,4). 

PQ = (4_ 2 )i + (2-3)j + (-l-5)k = 2i-j-6k 
PR = (3 - 2)1 + (6 - 3)j + (4 - 5)k = i + 3 j - k 

Area of triangle = £|PQXPR| = £ |(2i- j-6k) X (i + 3j-k)| 

i j k 
= il 2 -1 -6 
1 3-1 
= iV(19) 2 + (-4)2 + (7)2 = ■Jv'426 

TRIPLE PRODUCTS 

1.20. Show that A • (B x C) is in absolute value equal 
to the volume of a parallelepiped with sides 
A, B and C. 

Let n be a unit normal to parallelogram /, having 
the direction of B X C, and let h be the height of the 
terminal point of A above the parallelogram /. Fig. 1-26 

Volume of parallelepiped = (height fc)(area of parallelogram J) 

= (A.n)(|BXC|) 

= A»{|BXC|n} = A'(BXC) 

If A, B and C do not form a right-handed system, A • n < and the volume = | A • (B X C) | . 

1.21. (a) If A = Aii + A 2 j+Ask, B = Bd + B 2 j + Bsk, C = Cii + C 2 j + Ctk show that 

Ai A 2 Az 

A-(BxC) = Bi B 2 B 3 

C\ C 2 Ci 




16 



VECTORS, VELOCITY AND ACCELERATION 



[CHAP. 1 



(6) Give a geometric significance of the case where A • (B x C) = 0. 

i j k 

(a) A'(BXC) = A* B t B 2 B 3 

Cj C 2 C 3 

= (A i i + A 2 j + A 3 k) • [(B 2 C 3 - £ 3 C 2 )i + {B 3 C X - B X C 3 )\ + (B X C 2 - B 2 C t )k] 

A x A 2 A3 
= A 1 (B 2 C 3 -B 3 C 2 )+A 2 (B 3 C 1 -B 1 C 3 ) + A 3 {B X C 2 - B 2 C t ) = B x B 2 B 3 

C x C 2 c 3 

(b) By Problem 1.20 if A • (B X C) = then A, B and C are coplanar, i.e. are in the same plane, 
and conversely if A, B, C are coplanar then A • (B X C) = 0. 



1.22. Find the volume of a parallelepiped with sides A = 3i — j, B = j + 2k, C = i + 5j + 4k. 

3-10 
By Problems 1.20 and 1.21, volume of parallelepiped = |A»(BXC)| = 



12 
15 4 



= 1-201 = 20. 



1.23. If A = i + j, B = 2i-3j + k, C = 4j-3k, find (a)(AxB)xC, (b)Ax(BxC). 

i j k 

110 
2-3 1 



(a) A X B = 



= i - j - 5k. Then (A X B) X C = 



i j k 

1 -1 -5 
4-3 



= 23i + 3j + 4k. 



(6) B X C = 



i J k 

2-3 1 
4-3 



= 5i + 6j + 8k. Then A X (B X C) 



i 


j 


k 


1 


1 





5 


6 


8 



= 8i-8j + k. 



It follows that, in general, (A X B) X C ^ A X (B X C). 



DERIVATIVES AND INTEGRALS OF VECTORS 
1.24. If r = (i 8 + 2i)i-3<?- 2t j + 2sin5*k, find (a) ^, (b) 






d 2 r 



dt 2 



at t = 0. 



(a) ^ = ~(*s + 2t)i + ^(-3e- 2 *)j + ^(2 sin 5*)k = (3*2 + 2)i + 6e~2tj + 10 cos ht k 
At * = 0, dr/dt = 2i + 6j + 10k. 

(6) From (a), \dr/dt\ = V(2) 2 + (6) 2 + (10)2 = ^140 = 2\/35 at t = 0. 

(c) W = Jtilt) = ^<( 3 * 2 + 2 ) i + 6e ~ 2t J + 10cos5«k} = 6<i-12e-2tj-50sin5tk 

At t =t 0, d?r/dtf = — 12j. 

(d) From (c), Idtr/dP] =12 at t = 0. 



1.25. Prove that -f-(A-B) = A-^ + ^-B, where A and B are differentiate func- 



tions of u. 



du 



du tint, 



CHAP. 1] VECTORS, VELOCITY AND ACCELERATION 17 

„ L , , d ,. „, .. (A + AA) • (B + AB) - A'B 

Method 1. -j-(A»B) = lim * . 

du v am-*o Am 

A'AB + AA'B + AA'AB 

= lim ■ t 

Au-»0 AM 

/ AB AA AA \ . dB dA 

= ?™o{ A '^ + -^- B+ -^' AB ) = A S£ + 1S' B 

Method 2. Let A = A x i + A 2 j + A 3 k, B = Bji + B^ + £ 3 k. Then 

£< A " B > = ^A + aa + aw 



^ <Zi? 2 df? 3 \ , (dA x .dA 2 dA* \ 

A *-dir + A *~dV + A » *r; + {~dV Bl + ~dV B2 + ~du~ Bs ) 



. dB , dA 
dt* du 

d 2 
1.26. If <£(#, y, z) = z 2 2/s and A = Sx 2 yi + yz 2 \ - xzk, find ^-^ (<f>A) at the point (1, -2, -1). 

$A = {x 2 yz){Sx 2 yi + yz 2 \ - xzk) = 3x*y 2 zi + x 2 y 2 z 3 j - x 3 2/s 2 k 

T-(*A) = ^-(SajVrf + aV*^ -«■»«*) = Sx*y 2 i + 3x 2 y 2 z 2 j - 2x3yzk 

02 OZ 

3 2 _, 3 



(*A) = ^- (3a^i/ 2 i + Sx 2 y 2 z 2 j - 2x 3 wzk) = 6as 4 ]/i + 6»; 2 i/3 2 j - 2a; 3 *k 



dydz 

If « = 1, y = -2, z = -1, this becomes -12i - 12j + 2k. 

1.27. Evaluate f A(w)dw if A(u) = (3w 2 - l)i + (2i* - 3) j + {Qu 2 - 4tt)k. 

The given integral equals 

I {(3m 2 - l)i + (2m - 3)j + (6m 2 - 4M)k} du 



2 

u=l 



= (m 3 - M)i + (m 2 - 3m) j + (2m 3 - 2M 2 )k 

= {(8 - 2)i + (4 - 6)j + (16 - 8)k> - {(1 - Di + (1 - 3)j + (2 - 2)k} 
= 6i + 8k 

VELOCITY AND ACCELERATION 

1.28. A particle moves along a curve whose parametric equations are x = 3e~ 2t , y = 4 sin 3t, 

2 = 5 cos St where t is the time. 

(a) Find its velocity and acceleration at any time. 

(6) Find the magnitudes of the velocity and acceleration at £ = 0. 

(a) The position vector r of the particle is 

r = xi + yj + zk = 3e~ 2t i + 4 sin St j + 5 cos Bt k 

Then the velocity is 

v = dt/dt = -6e- 2t i + 12 cos Zt j - 15 sin St k 

and the acceleration is 

a = dv/dt = d 2 v/dt 2 = 12e- 2t i - 36 sin St j - 45 cos 3* k 

(6) At t = 0, v = dr/dt = -6i + 12j and a = d 2 r/dt 2 = 12i-45k. Then 
magnitude of velocity at t = is V(-6) 2 + (12) 2 = 6^5 
magnitude of acceleration at * = is V(12) 2 + (-45) 2 = 3 V241 



18 



VECTORS, VELOCITY AND ACCELERATION 



[CHAP. 1 



1.29. A particle travels so that its acceleration is given by 

a = 2e -t i + 5 cos t j - 3 sin t k 
If the particle is located at (1,-3,2) at time t = and is moving with a velocity 
given by 4i - 3j + 2k, find (a) the velocity and (b) the displacement of the particle 
at any time t>0. 

/ v dPr dv 

(°) a = -no = -jT = 2e-'i + 5 cos * j - 3 sin * k 



8 = W = ~dt = 2e_1 + 5cos * 
Integrating, v = J (2e"'i + 5 cos t j - 3 sin t k) dt 

= —2e~H + 5 sin t j + 3 cos t k + c t 
Since v = 4i-3j + 2k at t = 0, we have 

4i - 3j + 2k = -2i + 3k + c t or c t = 6i - 3j - k 
Then v = -2«-*i + 5 sin t j + 3 cos t k + 6i - 3j - k 

= (6 - 2e-*)i + (5 sin t - 3)j + (3 cos * - l)k 
(6) Replacing v by dr/dt in (I) and integrating, we have 

r = J [(6 - 2e-*)i + (5 sin * - 3)j + (3 cos t - l)k) dt 

= (6t + 2«~*)i - (5 cos * + 8«)j + (3 sin t - t)k + c 2 
Since the particle is located at (1,-3,2) at t = 0, we have r = i — 3 j + 2k at t = 0, so that 
i - 3j + 2k = 2i - 5j + c 2 or c 2 = -i + 2j + 2k 

Thus r = (6t + 2«~*-l)i + (2 - 5 cos * - 3«)j + (3 sin « - t + 2)k {2) 



(1) 



RELATIVE VELOCITY AND ACCELERATION 

1.30. An airplane moves in a northwesterly 
direction at 125 mi/hr relative to the 
ground, due to the fact that there is a 
westerly wind [i.e. from the west] of 
50 mi/hr relative to the ground. Deter- 
mine (a) graphically and (b) analyti- 
cally how fast and in what direction 
the plane would have traveled if there 
were no wind. 

(a) Graphically. 

Let W = wind velocity 

V a = velocity of plane 
with wind 

V(, = velocity of plane 
without wind. 

Then [see Pig. 1-27] V a = V b + W or 




Fig. 1-27 

v 6 = v a -w = V„ + (-W). 



V b has magnitude 6.5 units = 163 mi/hr and direction 33° north of west. 

(6) Analytically. 

Letting i and j be unit vectors in directions E and N respectively, we see from Fig. 1-27 
that 

V = - 125 cos 45° i + 125 sin 45° j and W = 50i 

Then V„ = V a - W = (- 125 cos 45° - 50)i + 125 sin 45° j = -138.391 + 88.39J. 

Thus the magnitude of V b is V(-138.39) 2 + (88.39) 2 = 164.2 mi/hr and the direction is 
tan-i 88.39/138.39 = tan~i .6387 = 32° 34' north of west. 



CHAP. 1] VECTORS, VELOCITY AND ACCELERATION 19 

1.31. Two particles have position vectors given by ri = 2ti - t 2 j + (St 2 - 4£)k and 
r 2 = (U 2 - 12* + 4)i + Fj - 3tk. Find (a) the relative velocity and (b) the relative 
acceleration of the second particle with respect to the first at the instant where t = 2. 

(a) The velocities of the particles at t-2 are respectively 



Vl = r t = 2i-2tj + (6t-4)k 
v 2 = f 2 = (lOt - 12)i + St 2 } - 3k 



= 2i - 4j + 8k 

t = 2 



= 8i + 12j - 3k 

t = 2 



Relative velocity of particle 2 with respect to particle 1 

= y 2 - Vl = (8i + 12j - 3k) - (2i - 4j + 8k) = 6i + 16j - Ilk 

(6) The. accelerations of the particles at t = 2 are respectively 



a i — v i = *i = ~2j + 6k 

a 2 = v 2 = r 2 = lOi + 6tj 



= -2j + 6k 

t = 2 

= 10i + 12j 

t = 2 



Relative acceleration of particle 2 with respect to particle 1 
= a 2 - a! = (lOi + 12j) - (-2j + 6k) = lOi + 14j - 6k 

TANGENTIAL AND NORMAL ACCELERATION 
1.32. Given a space curve C with position vector 

r = 3cos-2ti'+ 3 sin 2t j + (8*-4)k 

(a) Find a unit tangent vector T to the curve. 

(b) If r is the position vector of a particle moving on C at time t, verify in this 
case that v = vT. 

(a) A tangent vector to C is 

dr/dt = - 6 sin 2* i + 6 cos 2t j + 8k 

The magnitude of this vector is 



\dr/dt\ = ds/dt = V(- 6 sin 2t)* + (6 cos 2*)* + (8)* = 10 

Then a unit tangent vector to C is 

dr/dt dx/dt dt - 6 sin 2* i 4- 6 cos 2t j + 8k 



T = 



dx/dt ds/dt ds 10 

= - 1 sin 2t i + f cos 2t j + $ k 



(6) This follows at once from (a) since 

v = dr/dt = - 6 sin 2t i + 6 cos 2t j + 8k 

= (10)(- f sin 2* i + f cos2t j + $k) = vT 

Note that in this case the speed of the particle along the curve is constant. 

1.33. If T is a unit tangent vector to a space curve C, show that dT/ds is normal to T. 

Since T is a unit vector, we have T*T = 1. Then differentiating with respect to 8, we obtain 

T .f + *. T = 2T M = o or T.£= 
d8 da d8 da 

which states that dT/ds is normal, i.e. perpendicular, to T. 



20 VECTORS, VELOCITY AND ACCELERATION [CHAP. 1 

If N is a unit vector in the direction of dT/ds, we have 

dT/ds = kS 

and we call N the unit principal normal to C. The scalar k = \dT/ds\ is called the curvature, 
while R = 1/k is called the radius of curvature. 

1.34. Find the (a) curvature, (b) radius of curvature and (c) unit principal normal N to 
any point of the space curve of Problem 1.32. 

(a) From Problem 1.32, T = -f sin It i + § cos 2t j + |k. Then 

dT _ dT/dt _ (-6/5) cos 2t i - (6/5) sin 2t j 
ds ds/dt 10 



= — Jj cos 2t i — ^ sin It j 



Thus the curvature is c = 



= V(-^cos 2 *) 2 + (-^sin2«)2 = £ 



ds 

(b) Radius of curvature = R = 1/ K = 25/3 

(c) From (a), (6) and Problem 1.33, 

., 1 dT n dT _ x . . „ . 

N = KdF = ^-rfT = -cos2*i-sm2«j 

1.35. Show that the acceleration a of a particle which travels along a space curve with 
velocity v is given by , 9 

a = 3t T -+ff N 

where T is the unit tangent vector to the space curve, N is its unit principal normal 
and R is the radius of curvature. 

Velocity v = magnitude of v multiplied by unit tangent vector T, or 

v = vT 

Differentiating, a = % = * {vT) = ^ T + ^ 

dt dt dt dt 

But -^ = £?I ds_ ds = tjN 

dt ds dt KXN dt KVS * R 

ti.«« — ^ V m . /vfi\ dv _ , i;2 T 

Then a - * T + *(«-; = d* T + b n 

This shows that the component of the acceleration is dv/dt in a direction tangent to the path and 
v 2 /R in the direction of the principal normal to the path. The latter acceleration is often called 
the centripetal acceleration or briefly normal acceleration. 

CIRCULAR MOTION 

1.36. A particle moves so that its position vector is given by r = cos o>t i + sin U j where 
o> is a constant. Show that (a) the velocity v of the particle is perpendicular to r, 
(b) the acceleration a is directed toward the origin and has magnitude proportional 
to the distance from the origin, (c) r x v = a constant vector. 

/ \ dr 

di ~ ~ a sm ut * "*" " cos w * *' Then 

r • v = [cos at i + sin at j] • [— « sin at i + w cos at j] 

= (cos at)(—a sin at) + (sin at)(a cos at) = 

and r and v are perpendicular. 



CHAP. 1] 



VECTORS, VELOCITY AND ACCELERATION 



21 



(b) ^ = ^ = -a 2 cos at i - a 2 sin «t j = -w 2 [cos <ot i + sin at j] = -« 2 r 
7 at 2 at 

Then the acceleration is opposite to the direction of r, i.e. it is directed toward the origin. 

Its magnitude is proportional to |r| which is the distance from the origin. 

(c) r X v = [cos at i + sin at j] X [— a sin at i + a cos at j] 
i j k 

cos at sin at 

— a sin at a COS at 

Physically, the motion is that of a particle moving on the circumference of a circle with 
constant angular speed a. The acceleration, directed toward the center of the circle, is the 
centripetal acceleration. 



= «(cos 2 at + sin 2 at)k = «k, a constant vector. 



GRADIENT, DIVERGENCE AND CURL 

1.37. If = x 2 yz* and A = xzi-y 2 j + 2x 2 yk, find (a) V</>, (&) V'A, (c) V x A, (d) div 
(<j>A), (e) curl (<f>A). 

1 ' v \3« dy ' dz }* dx dy* dz 

= -^(x2yz3)i + £-(x*yz3)j + |^(as»y*»)k = 2<ci/z 3 i + ****j + 3x*y**k 

(6) VA = (£l + £j + £k)-<^-lrt + *«»*> 



(c) 



VXA = (^ i + ^ + ^ k ) X( ^ i " 2/2J + 



2x 2 yk) 



i j k 

3/fla; d/dy d/dz 
xz —y 2 2x^y 



= 2x 2 i + (x — 4*2/) j 
(d) div (*A) = V • (*A) = V • (x^yzH - x*yHH + 2* 4 i/ 2 z 3 k) 



+ f-(2a*2/ 2 z 3 ) 
oz 



(e) curl (*A) = V X (*A) = V X (x*yz*i - *WJ + 2* 4 i/ 2 z 3 k) 

i j k 

d/dx d/dy d/dz 

xSyz* -x 2 y*z* 2ar% 2 « 8 
= (4x*yz s + 3« 2 i/ 3 « 2 )i + (4«3j/2 3 - 8x 3 i/ 2 z 3 )j - (2xy s z* + o 3 « 4 )k 



1.38. (a) If A = (2xy + z?)i + (x 2 + 2y)j+{Sxz 2 -2)k, show that VXA 
(b) Find a scalar function <j> such that A = V«/>- 
i j k 

(a) VXA = d/dx d/dy d/dz = 

2xy + « 3 a 2 + 2y 3s« 2 - 2 



= 0. 



22 VECTORS, VELOCITY AND ACCELERATION [CHAP. 1 

(b) Method 1. If A = V<& = ^i + f2j + ^k then we must have 

ox dy dz 

(1) f£ = 2xy + * 3 (*) |£ = x2 + 2y (3) '& = 3xz2 - 2 

da; fly flz 

Integrating, we find 

U) 4> = sfiy + xtfl + F^x) (5) * = x 2 y + y 2 + F 2 (x,z) 

(6) <f» = xz*-2z + F 3 (x,y) 

Comparing these we must have F^y^) = y 2 — 2z, F 2 (x,z) = xz* — 2z, F z (x,y) = x*y + y* 
and so <f> = x 2 y + xz s + y 2 — 2z. 

Method 2. We have if A = v>, 

A-dr = ^i + ^j + ^k^.^i + ^j + ^k) 

= H dx + % dy + fz dz = d * 
an exact differential. For this case, 

d<f> = A'dr = (2xy + z*)dx + (x 2 + 2y)dy + (Sxz 2 -2)dz 

= [(2xy + z s ) dx + x 2 dy + Sxz 2 dz] + 2y dy - 2 dz 
= d(x 2 y + xz*) + d(y 2 ) + d(-2z) 
- d(x 2 y + xz* + y 2 - 2z) 
Then <p = x 2 y + xz s + y 2 — 2z. Note that an arbitrary constant can also be added to <f>. 



LINE INTEGRALS AND INDEPENDENCE OF THE PATH 

1.39. If A = (Sx 2 - 6yz)i + (2y + 3xz)j + (1 - 4xyz 2 )k, evaluate J A'dr from (0,0,0) to 
(1, 1, 1) along the following paths C: c 

(a) x = t, y = t 2 , z-i^. 

(b) the straight lines from (0, 0, 0) to (0, 0, 1), then to (0, 1, 1), and then to (1, 1, 1). 

(c) the straight line joining (0,0,0) and (1,1,1). 

f A • dr = f {(Sx 2 - 6yz)i + (2y + 3xz)j + (1 - 4xyz 2 )k} '(dxi + dyj + dz k) 
= I (Sx 2 -6yz)dx + (2y + Sxz)dy + (l-4xyz 2 )dz 

(a) If x = t, y = t 2 , z = «*, points (0, 0, 0) and (1, 1, 1) correspond to t = and t = 1 respectively. Then 

f A-dr = f {3*2 - 6(*2)(*3)} dt + {2*2 + 3(*)(*3)} d(t 2 ) + {1 - 4(*)(*2)(*3)2} d(*») 
J o J t=0 

f (3*2-6* 5 )d* + (4t? + 6t*)dt + (St 2 - 12* 11 ) dt = 2 
J t=o 

Another method. 

Along C, A = (3*2 - 6*5)i + (2*2 + 3*4)j + (i _ 4«9) k an d r = »i + yj + «k = *i + * 2 j + *3fc, 
dx = (i + 2<j + 3< 2 k) dt. Then 



J A'dr = J (3*2- 



6*5) dt + (4*8 + 6*5) dt + (St 2 - 12*") dt = 2 



(6) Along the straight line from (0, 0, 0) to (0, 0, 1), x — 0, y = 0, dx = 0, dy = while z varies from 
to 1. Then the integral over this part of the path is 

f {3(0) 2 -6(0)(z)}0 + {2(0) + 3(0)(«)}0 + {1 - 4(0)(0)(z 2 )} dz = f dz = 1 



CHAP. 1] 



VECTORS, VELOCITY AND ACCELERATION 



23 



Along the straight line from (0, 0, 1) to (0, 1, 1), x = 0, z = 1, dx = 0,dz = while y varies 
from to 1. Then the integral over this part of the path is 



2ydy = 1 



l r 1 

f {3(0) 2 - 6(y){l)}0 + {2i/ + 3(0)(1)} dy + {1 - 4(0)(i/)(l) 2 }0 = I 
J y =o ^y=o 

Along the straight line from (0, 1, 1) to (1, 1, 1), y = l,z = 1, dy = 0,dz = while x varies 
from to 1. Then the integral over this part of the path is 

C {3*2 - 6(1)(1)} dx + {2(1) + 3z(l)}0 + {1 - 4*(1)(1) 2 >0 = f (3* 2 - 6) dx = -5 



x=0 

Adding, I 

Jr. 



A • dr = 1 + 1-5 = -3. 



(c) Along the straight line joining (0,0,0) and (1,1,1) we have x = t, y = t, z = t. Then since 

dx = dy = dz = dt, 

( A'dr = f (3x 2 -6yz)dx + (2y + Sxz) dy + (1 - 4xyz 2 ) dz 

= f (3t 2 -6t*)dt + (2t + St 2 )dt + (l-4t*)d« 

f (2« + l-4t*)dt = 6/5 
Note that in this case the value of the integral depends on the particular path. 

1.40. If A = (2xy + z?)i + (x 2 + 2y)j + (3zz 2 -2)k show that (a) J A -dr is independent 

of the path C joining the points (1,-1,1) and (2,1,2) and (b) find its value. 

By Problem 1.38, V X A = or A • dr = d<p = d(x 2 y + xz* + y 2 - 2z). Then the integral is 
independent of the path aijd its value is 

J (2,1,2) /•(2.1.2) 

A'dr = I d(x*y + xz 3 + y 2 - 2c) 

ii.-i.i) *^(i,-i.D 



= x 2 y + xz* + |/2-22 



(2.1,2) 



(1,-1.1) 



= 18 



MISCELLANEOUS PROBLEMS 

1.41. Prove that if a and b are non-collinear, then xa + yb = implies x - y = 0. 

Suppose x ¥= 0. Then a?a + yb = implies xa = — yb or a = —(y/x)b, i.e. a and b must be parallel 
to the same line (collinear), contrary to hypothesis. Thus x = 0; then 2/b = 0, from which y = 0. 



1.42. Prove that the diagonals of a parallelogram bi- 
sect each other. 

Let ABCD be the given parallelogram with diagonals 
intersecting at P as shown in Fig. 1-28. 

Since BD + a = b, BD = b - a. Then BP = x(b - a). 

Since AC = a + b, AP = y(a + b). 

But AB = AP + PB = AP - BP, 

i.e. a = i/(a + b) — x(b — a) = (x + y)a + (y — x)b. 

Since a and b are non-collinear we have by Problem 
1.41, x + y = 1 and y — x = 0, i.e. x = y = % and P 
is the midpoint of both diagonals. 




24 



VECTORS, VELOCITY AND ACCELERATION 



[CHAP. 1 



1.43. Prove that for any vector A, 

(a) A = (A-i)i + (A-j)j + (A-k)k 

(b) A = A(cos a i + cos /? j + cos y k) 

where a, /?, y are the angles which A makes with i, j, k respectively and cos a, cos (3, 
cos y are called the direction cosines of A. 

(a) We have A = A t i + A 2 j + A 3 k. Then 

A«i = (A x i + A 2 j + A 3 k) • i = A x 
A»j = (Aji + Aaj + Agk)'! = A 2 
A'k = (Aii + AaJ + AgkJ-k = A 3 



Thus 



A = (A • i)i + (A • j)j + (A • k)k 



(b) 



A • i = |A| |i| cos a = A cos a 
A«j — |A||j|cos/? = Acos/3 

A»k = |A| |k| cos y = A cos y 
Then from part (a), 

A = (A • i)i + (A • j)j + (A • k)k = A(cosa i + cos/3 j + cosy k) 



1.44. Prove that V<£ is a vector perpendicular to the surface </>(#, 2/, 2) = c, where c is a 
constant. 

Let r = xi + j/j + zk be the position vector to any point P(x, y, z) on the surface. 
Then dv — dx i + dy j + dz k lies in the plane tangent to the surface at P. But 



* = ^ dx + ^ dy+ ^ dz 



or 
i.e. V# • dr = so that V^ is perpendicular to dr and therefore to the surface 



ff i + lf i+ lf k »- (&i + < "' i + &k) 







1.45. Find a unit normal to the surface 2# 2 + 4yz - 5z 2 = -10 at the point P(3,-l,2). 
By Problem 1.44, a vector normal to the surface is 

V(2ar2 + Ayz - 5z 2 ) = 4ai + 4zj + (Ay - 10z)k = 12i + 8j - 24k at (3,-1,2) 

12i + 8j - 24k 3i + 2j - 6k 



Then a unit normal to the surface at P is 



Another unit normal to the surface at P is 



V(12) 2 + (8) 2 + (-24)2 
3i + 2j - 6k 



1.46. A ladder AB of length a rests against a vertical wall OA [Fig. 1-29]. The foot B 
of the ladder is pulled away with constant speed vo. {a) Show that the midpoint 
of the ladder describes the arc of a circle of radius a/2 with center at O. (b) Find 
the velocity and speed of the midpoint of the ladder at the instant where B is distant 
b < a from the wall. 

(a) Let r be the position vector of midpoint M of AB. 
If angle OB A = o, we have 

OB = a cos e i, OA = a sin e j 
AB = OB — OA = a cos i — a sin j 

Then 

r = OA + AM = OA + £AB 

= a sin e j + \(a> cos e i — a sin e j) 

= ^a(cos 9 i + sin e j) 

Thus |r| = ^a, which is a circle of radius a/2 

with center at 0. Fig. 1-29 




CHAP. 1] 



VECTORS, VELOCITY AND ACCELERATION 



25 



(6) The velocity of the midpoint M is 

-jr = -jrtya(co& 8 i + sin e j)} = ^a(— sin e ei + cos 8 8j) 
where 8 = d8/dt. 

The velocity of the foot B of the ladder is 

vA = -r-(OB) = 37 (a cos 0i) = — a&m8 8i or a sin 8 8 = 
at at 

At the instant where B is distant 6 from the wall we have from (2), 



-v 



am 8 = 



Va 2 - 62 



8 = 



-v 



-n 



a a sin* y a 2 _ b 2 

Thus from (1) the required velocity of M at this instant is 



dr 1 / . 



y/cP-b* 



and its speed is av„/2 Va 2 — o 2 . 



(i) 



(«) 



1.47. Let (r, 0) represent the polar coordinates describing the position of a particle. If 
ri is a unit vector in the direction of the position vector r and $i is a unit vector 
perpendicular to r and in the direction of increasing 6 [see Fig. 1-30], show that 

(a) ri = cos 9 i 4- sin 6 j, 0i = — sin 6 i + cos j 
(6) i = cos ri — sin 0i, j = sin ri + cos 6 $i 



(a) If r is the position vector of the particle at any 
time t, then dr/dr is a vector tangent to the 
curve 8 = constant, i.e. a vector in the direc- 
tion of r (increasing r). A unit vector in this 
direction is thus given by 



_ 5r /Idrl 
dr/ \dr\ 



(1) 



Since 

r = xi + yj = rcosfli + rsintfj 
as seen from Fig. 1-30, we have 



3r . , . . \dr\ 

— = cos* i + sin j, — = 1 

dr \dr\ 



so that 




cos 8 i + sin 8 j 



(3) 



Fig. 1-30 



Similarly, Br/d8 is a vector tangent to the curve r = constant. A unit vector in this 
direction is thus given by 

Now from (2), 



dr ... . 3r 

— = — r sin 8i + rcosff j, — = r 

do * \de 



so that (-4) yields 



8 X = — sin 8 i + cos 8 j (5) 

(b) These results follow by solving the simultaneous equations (3) and (5) for i and j. 



26 VECTORS, VELOCITY AND ACCELERATION [CHAP. 1 

1.48. Prove that (a)r a = 00i (b)di = -0r u 
(a) Prom (S) of Problem 1.47 we have 

Tl dt dr dt de dt 



= (0)(f ) + (- sin e i + cos e })($) = e0 t 



(b) From (5) of Problem 1.47 we have 

_ d*i _ d*idr , ^ide_ 
01 ~ dt ~ dr dt de dt 

= (0)(r) + (— cos e i — sin e j)(e) = — er x 

1.49. Prove that in polar coordinates (a) the velocity is given by 

v = rri 4- r$6i 
and (b) the acceleration is given by 

a = ( r - r6 2 )ri + (r 6 + 2r9) 9 1 

(a) We have r = rr 1 so that 

dr dr , ^ r i ... . . 

v = Tt = Tt r * + r -dT = ^ + rr ^ = rr i + *"i 

by Problem 1.48(a). 

(6) From part (a) and Problem 1.48 we have 

dv d .* , • x 

a = Tt = de (rr ^ + r ^ l) 

= rr t + rr x + r0«x + rtf*! + rebi 

= Vr! + f(5#i) + ffoi + r'e'9 1 + (re^—eri) 

= (r — r$ 2 )r t + (r6>'+2r£)0 1 



Supplementary Problems 

VECTOR ALGEBRA 

1.50. Given any two vectors A and B, illustrate geometrically the equality 4A+3(B-A) = A + 3B. 

1.51. Given vectors A, B and C, construct the vectors (a) 2A - 3B + £C, (6) C - £A + JB. 

1.52. If A and B are any two non-zero vectors which do not have the same direction, prove that pA + qB 
is a vector lying in the plane determined by A and B. 

1.53. (a) Determine the vector having initial point (2,-1,3) and terminal point (3,2,-4). (6) Find the 
distance between the two points in (a). Ans. (a) i + 3j — 7k, (6) \/59 

1.54. A triangle has vertices at the points A(2,l,-1), £(-1,3,2), C(l,-2,1). Find the length of the 
median to the side AB. Ans. £\/66 

1.55. A man travels 25 miles northeast, 15 miles due east and 10 miles due south. By using an 
appropriate scale determine (a) graphically and (6) analytically how far and in what direction 
he is from his starting position. Ans. 33.6 miles, 13.2° north of east. 



CHAP. 1] VECTORS, VELOCITY AND ACCELERATION 27 

1.56. Find a unit vector in the direction of the resultant of vectors A = 2i — j + k, B = i + j + 2k, 
C = 3i-2j + 4k. Ans. ±(61-21 + 7^/^ 

THE DOT OR SCALAR PRODUCT 

1.57. Evaluate |(A + B) • (A-B)| if A = 2i-3j + 5k and B = 3i + j-2k. Ans. 24 

1.58. Find a so that 2i — 3j + 5k and 3i + aj — 2k are perpendicular. Ans. a = —4/3 

1.59. If A = 2i + j + k, B = i - 2j + 2k and C = 3i - 4j + 2k, find the projection of A + C in the 
direction of B. Ans. 17/3 

1.60. A triangle has vertices at A(2,3,l), B(-l, 1,2), C(l,-2,3). Find the acute angle which the 
median to side AC makes with side BC. Ans. cos -1 V91/14 

1.61. Prove the law of cosines for triangle ABC, i.e. c 2 = a 2 + b 2 — 2ab cos C. 

[Hint. Take the sides as A,B,C where C = A — B. Then use OC = (A — B) • (A — B).] 

1.62. Prove that the diagonals of a rhombus are perpendicular to each other. 

THE CROSS OR VECTOR PRODUCT 

1.63. If A = 2i-j + k and B = i + 2j-3k, find |(2A + B) X (A-2B)|. Ans. 2byfl 

1.64. Find a unit vector perpendicular to the plane of the vectors A = 3i — 2j + 4k and B = i + j — 2k. 

Ans. ±(2j + k)/V5 

1.65. Find the area of the triangle with vertices (2,-3,1), (1,-1,2), (-1,2,3). Ans. \yf% 

1.66. Find the shortest distance from the point (3,2,1) to the plane determined by (1,1,0), (3,-1,1), 
(-1,0,2). Ans. 2 

* /.ft t> 0.1.7 * • j. x • i a ns, • sin A sin B sin C 

1.67. Prove the law of sines for triangle ABC, i.e. = — r — = • 

a o c 

[Hint. Consider the sides to be A, B, C where A + B + C = and take the cross product of both 
sides with A and B respectively.] 

TRIPLE PRODUCTS 

1.68. If A = 2i + j - 3k, B = i - 2j + k and C = -i + j - 4k, find (a) A • (B X C), (b) C • (A X B), 
(c) A X (B X C), (d) (A X B) X C. Ans. (a) 20, (6) 20, (c) 8i - 19j - k, {d) 25i - 15j - 10k 

1.69. Prove that A • (B X C) = (A X B) • C, i.e. the dot and the cross can be interchanged. 

1.70. Find the volume of a parallelepiped whose edges are given by A = 2i + 3j — k, B = i — 2j + 2k, 
C = 3i-j-2k. Ans. 31 

1.71. Find the volume of the tetrahedron with vertices at (2, 1, 1), (1, — 1, 2), (0, 1, -1), (1„ -2, 1). 
Ans. 4/3 

1.72. Prove that (a) A-(BXC) = B«(CXA) =a C • (A X B), 

(b) A X (B X C) = B(A • C) - C(A • B). 

1.73. (a) Let r v r 2 , r 3 be position vectors to three points P 1 ,P 2 ,P3 respectively. Prove that the equation 
(r — rj) • [(r — r 2 ) X (r — r 3 )] = 0, where r = xi + yj + zk, represents an equation for the plane 
determined by P u P 2 and P 3 . (&) Find an equation for the plane passing through (2,-1,-2), 
(-1,2,-3), (4,1,0). Ans. (b) 2x + y-3z = 9 

DERIVATIVES AND INTEGRALS OF VECTORS 

1.74. Let A = 3fi - (t 2 + t)j + (t* - 2t 2 )k. Find (a) dA/dt and (6) cPA/dt 2 at t = 1. 
Ans. (a) 3i-3j-k, (6) -2j + 2k 

1.75. If r = a cos ut + b sin at, where a and b are any constant non-collinear vectors and « is a constant 
scalar, prove that (a) r X dr/dt = <o(a X b), (6) d 2 r/dt 2 + <o 2 r = 0. 



28 VECTORS, VELOCITY AND ACCELERATION [CHAP. 1 

1.76. If A = t\ — sin t k and B = cos t i + sin t j + k, find ^(A«B). Ans. — t sin * 

1.77. Prove that t-(AXB) = AX-r- + -r-XB where A and B are differentiate functions of u. 

du ' du du 

1.78. If A(m) = 4(tt - l)i - (2u + 3)j + Gu 2 k, evaluate (a) J A(u) du, (b) J (ui - 2k) • A(u) du. 
Ans. (a) 6i - 8j + 38k, (b) -28 2 * 

1.79. Find the vector B(u) such that d 2 B/du 2 = Gul - 48w 2 j + 12k where B = 2i - 3k and dB/du = 
i + 5j for u = 0. Ans. (u s + u + 2)i + (5u - 4u 4 )j + (Gu 2 - 3)k 



1.80. 



Xd?A dA 

A X -j-^ dt = A X -rr + c where c is a constant vector. 



1.81. If R = x 2 yi - 2y 2 zj + xy 2 z>k, find 



3 2 B fl 2 R 
aa; 2 3j/ 2 



at the point (2,1,-2). Ans. IGyfh 



3 2 

1.82. If A = x*i - y] + xzk and B = yi + xj-xyzk, find ^y(AXB) at the point (1,-1,2). 

Ans. — 4i + 8j 

VELOCITY AND ACCELERATION 

1.83. A particle moves along the space curve r = (t 2 + *)i + (8* - 2)j + (2t» - 4t 2 )k. Find the 
(a) velocity, (6) acceleration, (c) speed or magnitude of velocity and (d) magnitude of accelera- 
tion at time t = 2. Ans. (a) 5i + 3j + 8k, (6) 21 + 16k, (c) ly[2, (d) 2^ 

1.84. A particle moves along the space curve defined by x = e-* cost, y — e~ t sin t, z = e~ t . Find 
the magnitude of the (a) velocity and (6) acceleration at any time t. 

Ans. (a) V3~e _t , (6) VEe-* 

1 .85. The position vector of a particle is given at any time t by r = a cos at i + 6 sin ut j + ct 2 k. 
(a) Show that although the speed of the particle increases with time the magnitude of the 
acceleration is always constant. (&) Describe the motion of the particle geometrically. 

RELATIVE VELOCITY AND ACCELERATION 

1.86. The position vectors of two particles are given respectively by r x = ti — 1 2 \ + (2t + 3)k and 
r 2 = (2t - 3« 2 )i + 4tj - £%. Find (a) the relative velocity and (&) the relative acceleration of the 
second particle with respect to the first at t = l. Ans. (a) -5i + 6j-5k, (6) -6i + 2j-6k 

1.87. An automobile driver traveling northeast at 26 mi/hr notices that the wind appears to be coming 
from the northwest. When he drives southeast at 30 mi/hr the wind appears to be coming from 
60° south of west. Find the velocity of the wind relative to the ground. 

Ans. 52 mi/hr in a direction from 30° south of west 

1.88. A man in a boat on one side of a river wishes to reach a point directly opposite him on the other 
side of the river. Assuming that the width of the river is D and that the speeds of the boat and 
current are V and v < V respectively, show that (a) he should start his boat ups tream at an angle 
of sin- 1 (v/V) with the shore and (6) the time to cross the river is D/y/V 2 - v 2 . 

TANGENTIAL AND NORMAL ACCELERATION 

1.89. Show that the tangential and normal acceleration of a particle moving on a space curve are given 
by d 2 s/dt 2 and K (ds/dt) 2 where s is the arc length of the curve measured from some initial point 
and k is the curvature. 

1.90. Find the (a) unit tangent T, (6) principal normal N, (c) radius of curvature R and (d) curvature 
k to the space curve x = t, y = t 2 /2, z = t. 

Ans. (a) (i+tj + k)/v^T2, (&) (-ti + 2j - tk)/VW+4, (c) (t 2 + 2f' 2 lyf2, (d) V2/(* 2 + 2) 3 ' 2 



CHAP. 1] VECTORS, VELOCITY AND ACCELERATION 29 

1.91. A particle moves in such a way that its position vector at any time t is r = £i + |^ 2 j + ik. 
Find (a) the velocity, (6) the speed, (c) the acceleration, (d) the magnitude of the acceleration, 
(e) the magnitude of the tangential acceleration, (/) the magnitude of the normal acceleration. 
Ans. (a) i + *j + k, (6) y/W+2, (c) j, (d) 1, (e) t/V& + 2, (/) V2/V^T2 

1.92. Find the (a) tangential acceleration and (6) normal acceleration of a particle which moves on 
the ellipse r = a cos at i + b sin at j. 

M 2( a 2 _ fr2) sm at cos at a 2 ab 

An8. (a) , (6) . 

Va 2 sin 2 at + b 2 cos 2 u£ V a 2 sin 2 <ot + 6 2 cos 2 at 

CIRCULAR MOTION 

1.93. A particle moves in a circle of radius 20 cm. If its tangential speed is 40 cm/sec, find (a) its 
angular speed, (6) its angular acceleration, (c) its normal acceleration. 

Ans. (a) 2 radians/sec, (6) radians/sec 2 , (c) 80 cm/sec 2 

1.94. A particle moving on a circle of radius R has a constant angular acceleration a. If the particle 
starts from rest, show that after time t (a) its angular velocity is a — at, (b) the arc length 
covered is 8 = ^Rat 2 . 

1.95. A particle moves on a circle of radius R with constant angular speed « . At time t = it starts 
to slow down so that its angular acceleration is —a (or deceleration a). Show that (a) it comes to 
rest after a time u /a and (6) has travelled a distance Raf,/2a. 

1.96. If the particle in Problem 1.95 is travelling at 3600 revolutions per minute in a circle of radius 
100 cm and develops a constant deceleration of 5 radians/sec 2 , (a) how long will it be before it 
comes to rest and (6) what distance will it have travelled? Ans. (a) 75.4 sec, (6) 1.42 X 10 e cm 

GRADIENT, DIVERGENCE AND CURL 

1.97. If A = xzi+(2x 2 -y)j-yz 2 k and <f> = $x 2 y + yW, find (a) V^, (6) V»A and (c) VXA 
at the point (1,-1,1). Ans. (a) -6i + j + 3k, (6) 2, (c) -i + j + 4k 

1.98. If <p = xy + yz + zx and A = x 2 y\ + yH\ + z 2 xk, find (a)A«V^, (6)^V»A and (c) (V#) X A 
at the point (3, -1, 2). Ans. (a) 25, (b) 2, (c) 56i - 30j + 47k 

1.99. Prove that if U,V,A,B have continuous partial derivatives, then (o) V(17 + V) = V17+VV, 
(b) V • (A + B) = V • A + V • B, (c) V X (A + B) = V X A + V X B. 

1.100. Show that V X (r 2 r) = where r = xi + yj + zk and r = |r|. 

1.101. Prove that (a) div curl A = and (6) curl grad <j> — under suitable conditions on A and <p. 

1.102. If A = (2x 2 -yz)i + (y 2 -2xz)j + x 2 z*k and <p = x 2 y - Sxz 2 + 2xyz, show directly that 
div curl A = and curl grad $ — 0. 

1.103. If A = 3»z 2 i - yzj + (x + 2z)k, find curl curl A. Ans. -6xi + (6*-l)k 

1.104. (a) Prove that V X (V X A) = -V 2 A+ V(V • A), (b) Verify the result in (a) if A is as given in 
Problem 1.103. 

1.105. Prove: (a) V X (UA) = (VC7) X A+ U{V X A). (6) V • (AX B) = B« (V X A) - A' (V X B). 

LINE INTEGRALS AND INDEPENDENCE OF THE PATH 

1.106. If F = (3aj - 2y)i + (y + 2z)j - x 2 k, evaluate \ F > dr from (0, 0, 0) to (1, 1, 1), where C is a path 

consisting of: (a) the curve x = t, y = t 2 , z = £ 3 ; (6) a straight line joining these points; (c) the 
straight lines from (0,0,0) to (0,1,0), then to (0,1,1) and then to (1,1,1); (d) the curve x = z 2 , 
z = y 2 . Ans. (a) 23/15, (6) 5/3, (c) 0, (d) 13/30 



30 VECTORS, VELOCITY AND ACCELERATION [CHAP. 1 

1.107. Evaluate j A-dr where A = 3x 2 i -K (2xz — y) j + zk along (a) the straight line from (0,0,0) 

to (2,1,3), (b) the space curve x = 2t 2 , y — t, z — 4t 2 — t from t = to t = l, (c) the curve defined 
by x 2 = 4y, 3x* = 8z from x = to x = 2. Ans. (a) 16, (6) 14.2, (c) 16 

1.108. Find 4> F»dr where F = (a; — 3#)i + (y — 2x)i and C is the closed curve in the xy plane, 
x = 2 cos t, i/ = 3 sin t, z = from t = to t = 2ir. Ans. 6tt 

1.109. (a) If A = {Axy - Zx 2 z 2 )i + (4y + 2ar 2 )j + (1 - 2x s z)k, prove that I A'dr is independent of the 

curve C joining two given points. (6) Evaluate the integral in (a) if C is the curve from the 
points (1,-1,1) to (2,-2,-1). Ans. (b) -19 

1.110. Determine whether I A • dr is independent of the path C joining any two points if (a) A = 2xyzi + 

J c 
x 2 zj + x 2 yk, (b) 2xzi + (x 2 — y)j + (2z — x 2 )k. In the case where it is independent of the path, 
determine <j> such that A = V#. 
Ans. (a) Independent of path, <f> = x 2 yz + c; (b) dependent on path 

1.111. Evaluate (J) E • dr where E = rr. Ans. 

MISCELLANEOUS PROBLEMS 

1.112. If A X B = 8i - 14j + k and A + B = 5i + 3j + 2k, find A and B. 
Ans. A = 2i + j-2k, B = 3i + 2j + 4k 

1.113. Let l^m^n! and Z 2 >™2« w 2 be direction cosines of two vectors. Show that the angle between them 
is such that cos e = IJ2 + m 1 m 2 + n 1 % 2 « 

1.114. Prove that the line joining the midpoints of two sides of a triangle is parallel to the third side 
and has half its length. 

1.115. Prove that (A X B) 2 + (A • B) 2 = A 2 B 2 . 

1.116. If A, B and C are non-coplanar vectors [vectors which do not all lie in the same plane] and 

x t A + y{B + «]C = x 2 A + y 2 B + z 2 C, prove that necessarily x x — x 2 , y\ — y%, z x — z 2 . 

1.117. Let ABCD be any quadrilateral and points P, Q, R and S the midpoints of successive sides. Prove 
that (a) PQRS is a parallelogram, (6) the perimeter of PQRS is equal to the sum of the lengths 
of the diagonals of ABCD. 

1.118. Prove that an angle inscribed in a semicircle is a right angle. 

1.119. Find a unit normal to the surface x 2 y - 2xz + 2y 2 z^ = 10 at the point (2, 1, -1). 
Ans. ±(Zi + 4}-6k)/^ 

- <„» ^ m ^ . dA . dA 

1.120. Prove that A • -^ = A -^ . 

1.121. If A(w) is a differentiate function of u and |A(te)| = 1, prove that dA/du is perpendicular to A. 

1.122. Prove V-(*A) = (V*) • A + <f>(V • A). 

1.123. If A X B = A X C, does B = C necessarily? Explain. 

1.124. A ship is traveling northeast at 15 miles per hour. A man on this ship observes that another ship 
located 5 miles west seems to be traveling south at 5 miles per hour, (a) What is the actual 
velocity of this ship? (6) At what distance will the two ships be closest together? 

1.125. Prove that (A X B) • (C X D) + (B X C) • (A X D) + (C X A) • (B X D) = 0. 

1.126. Solve the equation d 2 r/dt 2 = -gk where g is a constant, given that r = 0, dr/dt = vjs. at t = 0. 
Ans. r = (v t — ^gt 2 )k 



CHAP. 1] VECTORS, VELOCITY AND ACCELERATION 31 

1.127. If <t> = (« 2 + 2/ 2 + 2 2 )- 1/2 , show that V 2 ^ = V • (Vtf>) = at all points except (0,0,0). 

1.128. The muzzle velocity of a gun is 60 mi/hr. How long does it take for a bullet to travel through the 
gun barrel which is 2.2 ft long, assuming that the bullet is uniformly accelerated? Ans. .05 sec 

1.129. A 25 foot ladder AB rests against a vertical wall OA as in Fig. 1-29, page 24. If the foot of the 
ladder B is pulled away from the wall at 12 ft/sec, find (a) the velocity and (6) the acceleration 
of the top of the ladder A at the instant where B is 15 ft from the wall. 

Ans. (a) 9 ft/sec downward, (6) 11.25 ft/sec 2 downward 

1.130. Prove that (a)|A + B|^ |A| + |B|, (6) |A+B + C| ^ |A| + |B| + |C|. Give a possible geometric 
interpretation. 

1.131. A train starts from rest with uniform acceleration. After 10 seconds it has a speed of 20 mi/hr. 
(a) How far has it traveled from its starting point after 15 seconds and (6) what will be its speed 
in mi/hr? Ans. (a) 330 ft, (6) 30 mi/hr 

1.132. Prove that the magnitude of the acceleration of a particle moving on a space curve is 

y/(dvldt)* + vVR 2 
where v is the tangential speed and R is the radius of curvature. 

1.133. If T is the unit tangent vector to a curve C and A is a vector field, prove that 

JA'dr = \ A'Tds 
c J c 

where s is the arc length parameter. 

1.134. If A = (2x - y + 4)i + (5y + Sx - 6)j, evaluate 4> A'dr around a triangle with vertices at 
(0,0,0), (3,0,0), (3,2,0). Ans. 12 J 

1.135. An automobile driver starts at point A of a highway and stops at point B after traveling the 
distance D in time T. During the course of the trip he travels at a maximum speed V. Assuming 
that the acceleration is constant both at the beginning and end of the trip, show that the time 
during which he travels at the maximum speed is given by 2D/V — T. 

1.136. Prove that the medians of a triangle (a) can form a triangle, (b) meet in a point which divides the 
length of each median in the ratio two to one. 

1.137. If a particle has velocity v and acceleration a along a space curve, prove that the radius of 
curvature of its path is given numerically by 

R = ^-r 

|vXa| 

1.138. Prove that the area of a triangle formed by vectors A, B and Cis £|AXB + BXC + CXA|. 

1.139. (a) Prove that the equation AXX = B can be solved for X if and only if A»B = and A^0. 
(6) Show that one solution is X = B X A/A 2 , (c) Can you find the general solution? 

Ans. (c) X = B x A/A 2 + XA where X is any scalar. 

1.140. Find all vectors X such that A • X = p. 

Ans. X = pA/A 2 + V X A where V is an arbitrary vector. 

1.141. Through any point inside a triangle three lines are constructed parallel respectively to each of 
the three sides of the triangle and terminating in the other two sides. Prove that the sum of the 
ratios of the lengths of these lines to the corresponding sides is 2. 

1.142. If T, N and B = T X N are the unit tangent vector, unit principal normal and unit binormal to 
a space curve r = r(u), assumed differentiate, prove that 

dT dB *t < * n -«i' 

Is- = kN ' * = ~ tN ' Is- - tB ~ kT 
These are called the Frenet-Serret formulas. In these formulas k is called the curvature, r is the 
torsion and their reciprocals R = 1/k, o = Vt are called the radius of curvature and radius of 
torsion. 



32 



VECTORS, VELOCITY AND ACCELERATION 



[CHAP. 1 



1.143. In Fig. 1-31, AB is a piston rod of length I. If A moves along horizontal line CD while B moves 
with constant angular speed w around the circle of radius a with center at O, find (a) the velocity 
and (6) the acceleration of A. 

P 

P 




l-S 



l / 




Fig. 1-31 



Fig. 1-32 



1.144. A boat leaves point P [see Fig. 1-32] on one side of a river bank and travels with constant velocity 
V in a direction toward point Q on the other side of the river directly opposite P and distance D 
from it. If r is the instantaneous distance from Q to the boat, is the angle between r and PQ, 
and the river travels with speed v, prove that the path of the boat is given by 

_ D sec 6 

~ (sec e + tan e) v/v 

1.145. If v = V in Problem 1.144, prove that the path is an arc of a parabola. 

L146. (a) Prove that in cylindrical coordinates (p,<j>,z) [see Fig. 1-33] the position vector is 

r = p cos <t> i + p sin <f> j + zk 
(6) Express the velocity in cylindrical coordinates, 
(c) Express the acceleration in cylindrical coordinates. 

Ans. (6) v = ppx 4- p$$\ + «k 

(c) a = (p- P* 2 )pi + (p'4+ 2p£)tfi + *k 







Illlplllilplllliplim 






2 


* 


p'v. 






Cylindrical coordinates 
Fig. 1-33 



Spherical coordinates 
Fig. 1-34 



1.147. (a) Prove that in spherical coordinates (r, $, <j>) [see Fig. 1-34] the position vector is 

r = r sin e cos <f> i + r sin 9 sin # j + r cos k 

(6) Express the velocity in spherical coordinates, 
(c) Express the acceleration in spherical coordinates. 

Ana. (6) v = rt x + rb§ x + rj> sin $ f t 

(c) a = (V - rff 2 — r£ 2 sin 2 tf)r x + (2re + rV — r^ 2 sin cos *)«! 
+ (2re$ + 2r# sin 6 + r'<p sin 0)# x 

1.148. Show that if a particle moves in the xy plane the results of Problems 1.146 and 1.147 reduce to 
those of Problem 1.49. 



WORK, ENERGY and MOMENTUM 



NEWTON'S LAWS 

The following three laws of motion given by Sir Isaac Newton are considered the 
axioms of mechanics: 

1. Every particle persists in a state of rest or of uniform motion in a straight line 
(i.e. with constant velocity) unless acted upon by a force. 

2. If F is the (external) force acting on a particle of mass m which as a consequence is 
moving with velocity v, then 

F = 5<""> = % « 

where p = mv is called the momentum. If m is independent of time t this becomes 

dx 
F = m-jT = ma (#) 

where a is the acceleration of the particle. 

3. If particle 1 acts on particle 2 with a force Fi 2 in a direction along the line joining 
the particles, while particle 2 acts on particle 1 with a force F 2 i, then F 2 i = — F 12 . In 
other words, to every action there is an equal and opposite reaction. 

DEFINITIONS OF FORCE AND MASS 

The concepts of force and mass used in the above axioms are as yet undefined, although 
intuitively we have some idea of mass as a measure of the "quantity of matter in an object" 
and force as a measure of the "push or pull on an object". We can however use the above 
axioms to develop definitions [see Problem 2.28, page 49]. 

UNITS OF FORCE AND MASS 

Standard units of mass are the gram (gm) in the cgs (centimeter-gram-second) system, 
kilogram (kg) in the mks (meter-kilogram-second) system and pound (lb) in the fps (foot- 
pound-second) system. Standard units of force in these systems are the dyne, newton (nt) 
and poundal (pdl) respectively. A dyne is that force which will give a 1 gm mass an accelera- 
tion of 1 cm/sec 2 . A newton is that force which will give a 1 kg mass an acceleration of 
1 m/sec 2 . A poundal is that force which will give a 1 lb mass an acceleration of 1 ft/sec 2 . 
For relationships among these units see Appendix A, page 341. 

INERTIAL FRAMES OF REFERENCE. ABSOLUTE MOTION 

It must be emphasized that Newton's laws are postulated under the assumption that 
all measurements or observations are taken with respect to a coordinate system or frame 

33 



34 



NEWTON'S LAWS OF MOTION. WORK, ENERGY AND MOMENTUM 



[CHAP. 2 



of reference which is fixed in space, i.e. is absolutely at rest. This is the so-called assump- 
tion that space or motion is absolute. It is quite clear, however, that a particle can be 
at rest or in uniform motion in a straight line with respect to one frame of reference and 
be traveling in a curve and accelerating with respect to another frame of reference. 

We can show that if Newton's laws hold in one frame of reference they also hold in 
any other frame of reference which is moving at constant velocity relative to it [see 
Problem 2.3]. All such frames of reference are called inertial frames of reference or 
Newtonian frames of reference. To all observers in such inertial systems the force 
acting on a particle will be the same, i.e. it will be invariant. This is sometimes called 
the classical principle of relativity. 

The earth is not exactly an inertial system, but for many practical purposes can be 
considered as one so long as motion takes place with speeds which are not too large. For 
non-inertial systems we use the methods of Chapter 6. For speeds comparable with the 
speed of light (186,000 mi/sec), Newton's laws of mechanics must be replaced by Einstein's 
laws of relativity or relativistic mechanics. 



WORK 

If a force F acting on a particle gives it a 
displacement dr, then the work done by the 
force on the particle is defined as 



dW = F-dr 



(3) 



since only the component of F in the direction 
of dr is effective in producing the motion. 

The total work done by a force field (vector 
field) F in moving the particle from point Pi 
to point P 2 along the curve C of Fig. 2-1 is 
given by the line integral [see Chap. 1, page 9]. 




Fig. 2-1 



W = 



F-dr = I F-dr = \ F 
c *yPi "n 



dr 



(*) 



where ri and r 2 are the position vectors of Pi and P 2 respectively. 



POWER 

The time rate of doing work on a particle is often called the instantaneous power, or 
briefly the power, applied to the particle. Using the symbols W and <P for work and 
power respectively we have . 

r dt w 

If F is the force acting on a particle and v is the velocity of the particle, then we have 

<P = Fv (6) 



KINETIC ENERGY 

Suppose that the above particle has constant mass and that at times U and U it is 
located at Pi and P 2 [Fig. 2-1] and moving with velocities vi = dn/dt and v 2 = drjdt 
respectively. Then we can prove the following [see Problem 2.8]. 



CHAP. 2] NEWTON'S LAWS OF MOTION WORK, ENERGY AND MOMENTUM 35 

Theorem 2 J. The total work done in moving the particle along C from Pi to P 2 is 
given by 

W = C Fdr = ^m(v\-v\) (7) 

If we call the quantity T = \mv^ (8) 

the kinetic energy of the particle, then Theorem 2.1 is equivalent to the statement 

Total Work done from Pi to P2 along C 

= Kinetic energy at P2 — Kinetic energy at Pi 

or, in symbols, W = T 2 -Ti (10) 

where Ti — ^mv 2 v T 2 — \mv\. ^ 

I 

CONSERVATIVE FORCE FIELDS 

Suppose there exists a scalar function V such that F = - vV. Then we can prove the 
following [see Problem 2.15]. 

Theorem 2.2. The total work done in moving the particle along C from Pi to P 2 is 

W = f p 2 F-dr = V(Pi) - V(P 2 ) (11) 

In such case the work done is independent of the path C joining points Pi and P 2 . If the 
work done by a force field in moving a particle from one point to another point is 
independent of the path joining the points, then the force field is said to be conservative. 

The following theorems are valid. 

Theorem 23. A force field F is conservative if and only if there exists a continuously 
differentiable scalar field V such that F=-vF or, equivalently, if and only if 

V X F = curl F = identically (12) 

Theorem 2.4. A continuously differentiable force field F is conservative if and only 
if for any closed non-intersecting curve C (simple closed curve) 

i F-dr = (13) 

i.e. the total work done in moving a particle around any closed path is zero. 

POTENTIAL ENERGY OR POTENTIAL 

The scalar V such that F=-yF is called the potential energy, also called the scalar 
potential or briefly the potential, of the particle in the conservative force field F. In such 
case equation (11) of Theorem 2.2 can be written 

Total Work done from Pi to P 2 along C 

= Potential energy at Pi - Potential energy at P 2 
or, in symbols, W = Vi - V 2 (15) 

where Vi = V(Pi), V 2 = V(P 2 ). 



36 



NEWTON'S LAWS OF MOTIQN. WORK, ENERGY AND MOMENTUM [CHAP. 2 



It should be noted that the potential is defined within an arbitrary additive constant. 
We can express the potential as 

V i= - f F-dr (16) 

where we suppose that V = when r = r . 



CONSERVATION OF ENERGY 

For a conservative force field we have from equations (10) and (15), 

T 2 -Ti = Vi- V a or T1 + V1 = T 2 + V 2 
which can also be written \mv\ + Vi — \mv\ + V 2 



(17) 
(18) 



The quantity E = T + V, which is the sum of the kinetic energy and potential energy, is 
called the total energy. From (18) we s^ee that the total energy at Pi is the same as the 
total energy at P 2 . We can state our results in the following 

Theorem 2.5. In a conservative fofce field the total energy [i.e. the sum of kinetic 
energy and potential energy] is a constant. In symbols, T + V = constant = E. 

This theorem is often called the principle of conservation of energy. 



IMPULSE 

Suppose that in Fig. 2-1 the particle is located at Pi and P2 at times £1 and U where it 
has velocities vi and v 2 respectively. Thb time integral of the force F given by 



r 



Ydt 



(19) 



is called the impulse of the force F. The following theorem can be proved [see Problem 2.18]. 
Theorem 2.6. The impulse is equal to the change in momentum; or, in symbols, 



Xt 2 
F 



dt = m\ 2 — mvj 



Pi 



(20) 



The theorem is true even when the mass is variable and the force is non-conservative. 



TORQUE AND ANGULAR MOMENTUM 

If a particle with position vector r moves in a 
force field F [Fig. 2-2], we define 



A = rxF 



(21) 



as the torque or moment of the force F about 0. 
The magnitude of A is a measure of the j'turning 
effect" produced on the particle by the fcjrce. We 
can prove the following [see Problem 2.5J50] 



Theorem 2.7. 



rxF = -3r{m(rXv)} 



(22) 




Fig. 2-2 



CHAP. 2] NEWTON'S LAWS OF MOTION. WORK, ENERGY AND MOMENTUM 37 

The quantity 

12 = m(rxv) = rxp (23) 

is called the angular momentum or moment of momentum about 0. In words the theorem 
states that the torque acting on a particle equals the time rate of change in its angular 
momentum, i.e., 

- = w w 

This theorem is true even if the mass m is variable or the force non-conservative. 

CONSERVATION OF MOMENTUM 

If we let F = in Newton's second law, we find 

-jT-(rav) = or mv = constant (25) 

This leads to the following 

Theorem 2.8. If the net external force acting on a particle is zero, its momentum 
will remain unchanged. 

This theorem is often called the principle of conservation of momentum. For the case 
of constant mass it is equivalent to Newton's first law. 



CONSERVATION OF ANGULAR MOMENTUM 

If we let A = in (2U) f we find 
j 
-jr{m(rXv)} = or m(rXv) = constant (26) 

This leads to the following 

Theorem 2J9. If the net external torque acting on a particle is zero, the angular 
momentum will remain unchanged. 

This theorem is often called the principle of conservation of angular momentum. 



NON-CONSERVATIVE FORCES 

If there is no scalar function V such that F = - yV [or, equivalently, if V x F *■ 0], 
then F is called a non-conservative force field. The results (7), (20) and (2U) above hold 
for all types of force fields, conservative or not. However, (11) and (17) or (18) hold only 
for conservative force fields. 



STATICS OR EQUILIBRIUM OF A PARTICLE 

An important special case of motion of a particle occurs when the particle is, or appears 
to be, at rest or in equilibrium with respect to an inertial coordinate system or frame of 
reference. A necessary and sufficient condition for this is, from Newton's second law, that 

F = (27) 

i.e. the net (external) force acting on the particle be zero. 



38 NEWTON'S LAWS OF MOTION. WORK, ENERGY AND MOMENTUM [CHAP. 2 

If the force field is conservative with potential V, then a necessary and sufficient 
condition for a particle to be in equilibrium at a point is that 

F = . dV = dV = §V = 

at the point. ' ^ Bx dy dz 

STABILITY OF EQUILIBRIUM 

If a particle which is displaced slightly from an equilibrium point P tends to return 
to P, then we call P a point of stability or stable point and the equilibrium is said to be 
stable. Otherwise we say that the point is one of instability and the equilibrium is 
unstable. The following theorem is fundamental. 

Theorem 2.10. A necessary and sufficient condition that an equilibrium point be one 
of stability is that the potential V at the point be a minimum. 



Solved Problems 

NEWTON'S LAWS 

2.1. Due to a force field, a particle of mass 5 units moves along a space curve whose 
position vector is given as a function of time t by 

r = (2t s + t)i + (St 4 -t 2 + 8) j - 12t 2 k 

Find (a) the velocity, (b) the momentum, (c) the acceleration and (d) the force 
field at any time t . 

dr 

(a) Velocity = v = ^ = (6* 2 + l)i + (12*3 - 2*)j - 24tk 

(6) Momentum = p = mv = 5v = (30*2 + 5)j + ( 60 f3 _ io*)j - 120* 

(c) Acceleration = a = ^ = -^ - 12ti + (36*2 _ 2 )j - 24k 

(d) Force = F = j^ = m^ = 60ti + (180t 2 - 10) j - 120k 

2.2. A particle of mass m moves in the xy plane so that its position vector is 

r = a cos (d i + b sin <*>t j 

where a, b and «> are positive constants and a> b. (a) Show that the particle moves 
in an ellipse, (b) Show that the force acting on the particle is always directed 
toward the origin. 

(a) The position vector is 

r = xi + y] = a cos at i + b sin at j 

and so x = a cos at, y = b sin at which are 
the parametric equations of an ellipse having 
semi-major and semi-minor axes of lengths a 
and b respectively [see Fig. 2-3]. 
Since 

(*/o) a + (y/b) 2 = cos 2 at + sin 2 at - 1 
the ellipse is also given by as 2 /a 2 + y 2 /b 2 = 1. Fig. 2-3 




CHAP. 2] 



NEWTON'S LAWS OF MOTION. WORK, ENERGY AND MOMENTUM 



39 



(6) Assuming the particle has constant mass m, the force acting on it is 

d 2 



„ dv d 2 r 

F = m Tt = m w 



m -r-£ [(a cos ut)i + (b sin «t)j] 

= m[— w 2 a cos wt i — w 2 6 sin ut j] 

— —mw 2 [a cos ut i + b sin ut j] = — mw 2 r 

which shows that the force is always directed toward the origin. 



2.3. Two observers O and 0', fixed relative to 
two coordinate systems Oxyz and O'x'y'z' 
respectively, observe the motion of a par- 
ticle P in space [see Fig. 2-4]. Show that 
to both observers the particle appears to 
have the same force acting on it if and 
only if the coordinate systems are moving 
at constant velocity relative to each other. 

Let the position vectors of the particle in the 
Oxyz and O'x'y'z' coordinate systems be r and r' 
respectively and let the position vector of 0' 
with respect to O be R = r — r\ 




Fig. 2-4 



Relative to observers O and O' the forces acting on P according to Newton's laws are given 
respectively by d 2 r < p r , 

F = m w' F ' = m W 

The difference in observed forces is 

F ~ F = m ^ (r " r) = m l# 



and this will be zero if and only if 

d 2 R 
dt 2 



= 



dR 

dt 



= constant 



i.e. the coordinate systems are moving at constant velocity relative to each other. Such coordinate 
systems are called inertial coordinate systems. 

The result is sometimes called the classical principle of relativity. 



2.4. A particle of mass 2 moves in a force field depending on time t given by 

F = 24t 2 i + (36£-16)j - 12tk 

Assuming that at t = the particle is located at r = 3i - j + 4k and has velocity 
vo = 6i + 15j - 8k, find (a) the velocity and (b) the position at any time t. 

(a) By Newton's second law, 

2dv/dt = 24t 2 i + (36t-16)j - 12tk 

or 

dv/dt = 12tH + (18t-8)j - 6tk 

Integrating with respect to t and calling c x the constant of integration, we have 

v = UH + (9£ 2 -8t)j - 3t 2 k + c t 

Since v = v = 6i + 15j - 8k at t = 0, we have c x = 6i + 15j - 8k and so 
v = (4t 3 + 6)i + (9t 2 -Bt+ 15)j - (3* 2 + 8)k 



40 NEWTON'S LAWS OF MOTION. WORK, ENERGY AND MOMENTUM [CHAP. 2 

(6) Since v = dr/dt, we have by part (a) 

dr 

^ = (4^3 + 6 )i + (9*2 _ 8t + 16) j _ (3t 2 + 8 ) k 

Integrating with respect to t and calling c 2 the constant of integration, 
r = (t 4 + 6«)i + (3t 3 - 4«2 + I5t)j - (tf* + 8t)k + c 2 

Since r = r = 3i — j + 4k at t = 0, we have c 2 = 3i — j + 4k and so 

r = (t* + 6t + 3)i + (3*3 - 4* 2 + 15t - l)j + (4 - & - St)k 

2.5. A constant force F acting on a particle of mass m changes the velocity from vi to V2 
in time t. 

(a) Prove that F = m(v 2 — vi)/t. 

(b) Does the result in (a) hold if the force is variable? Explain. 

(a) By Newton's second law, , , _ 

dv _ dv F /<v 

m-j- = F or -77 = — (1) 

dt dt m y ' 

Then if F and m are constants we have on integrating, 

v = (F/m)t + c 2 

At t = 0, v = v x so that c x = Vj i.e. 

v = (F/m)t + vj (2) 

At t = r, v = v 2 so that v 2 = (F/m)r + v t 

i.e. F = m(v 2 — vi)/t (S) 

Another method. 

Write (1) as mdv = Fdt. Then since v=v! at t = and v = v 2 at t — r, we have 



I md\ — \ Ydt or m(v 2 — v^ = Ft 



vi 

which yields the required result. 

(b) No, the result does not hold in general if F is not a constant, since in such case we would not 
obtain the result of integration achieved in (a). 

2.6. Find the constant force in the (a) cgs system and (b) mks system needed to 
accelerate a mass of 10,000 gm moving along a straight line from a speed of 
54 km/hr to 108 km/hr in 5 minutes. 

Assume the motion to be in the direction of the positive x axis. Then if v x and v 2 are the 
velocities, we have from the given data V! = 54i km/hr, v 2 = 108i km/hr, m = 10,000 gm, 
t = 5 min. 

(a) In the cgs system 

m = 10 4 gm, v x = 54i km/hr = 1.5 X lQ3i cm/sec, v 2 = 3.0 X 103i cm/sec, t = 300 sec 

/v 2 — v t \ /1.5 X 10 3 i cm/sec \ 

Then F = ma = m ( . ) = (10 4 gm) ) 

V * / V 3 X 102 sec / 

= 0.5 X 10 5 i gm cm/sec 2 = 5 X 10 4 i dynes 
Thus the magnitude of the force is 50,000 dynes in the direction of the positive x axis. 

(6) In the mks system 

m = 10 kg, vj = 54i km/hr = 15i m/sec, v 2 = 30i m/sec, t = 300 sec 



CHAP. 2] NEWTON'S LAWS OF MOTION. WORK, ENERGY AND MOMENTUM 



41 



Then 



F = wa = m 



v 2 - Vl 



= (10 kg) 



15i m/sec 
300 sec 



= 0.5i kg m/sec 2 = 0.5i newtons 



Thus the magnitude is 0.5 newtons in the positive x direction. This result could also have been 
obtained from part (a) on noting that 1 newton = 10* dynes or 1 dyne = 10 newtons. 

In this simple problem the unit vector i is sometimes omitted, it being understood that the 
force F will have the direction of the positive * axis. However, it is good practice to work this 
and similar problems with the unit vector present so as to emphasize the vector character of force, 
velocity, etc This is especially important in cases where velocities may change their directions. 
See, for example, Problem 2.46, page 56. 



2.7. What constant force is needed to bring a 2000 lb mass moving at a speed of 60 mi/hr 
to rest in 4 seconds? 

We shall assume that the motion takes place in a straight line which we choose as the positive 
direction of the x axis. Then using the English absolute system of units, we have 

m = 2000 1b, v x = 60i mi/hr = 88i ft/sec, v 2 = Oi ft/sec, t = 4 sec 



Then 



F = wa = m 



v 2 - v x 



'-88i ft/sec \ 

= (2000 lb) ( ) 

4 sec / 



= -4.4 X 10 4 i ft lb/sec 2 = -4.4 X 10 4 i poundals 



Thus the force has magnitude 4.4 X 10* poundals in the negative x direction, i.e. in a direction 
opposite to the motion. This is of course to be expected. 



WORK, POWER, AND KINETIC ENERGY 

2.8. A particle of constant mass m moves in space under the influence of a force field F 
Assuming that at times U and U the velocity is vi and v 2 respectively, prove that 
the work done is the change in kinetic energy, i.e., 

tt 



X 



Work done 



- £•■*• - X. 



F*\dt 



•dv 



= im f * d(v • v) = £mv 2 * = %mv\ - -Jmv 2 



2.9. Find the work done in moving an object along a 
vector r = 3i + 2j-5k if the applied force is 
F = 2i - j - k. Refer to Fig. 2-5. 

Work done = (magnitude of force in direction 
of motion) (distance moved) 

= (Fcostf)(r) = Ft 

= (2i-j-k)-(3i + 2j-5k) 

= 6-2 + 5 = 9 




r 
Fig. 2-5 



42 NEWTON'S LAWS OF MOTION. WORK, ENERGY AND MOMENTUM [CHAP. 2 

2.10. Referring to Problem 2.2, (a) find the kinetic energy of the particle at points A 
and B, (b) find the work done by the force field in moving the particle from A to B, 
(c) illustrate the result of Problem 2.8 in this case and (d) show that the total work 
done by the field in moving the particle once around the ellipse is zero. 

(a) Velocity = v == dr/dt — — ua sin ut i + ub cos at j. 

Kinetic energy = |w 2 = ^m(u 2 a 2 sin 2 ut + u 2 b 2 cos 2 ut). 

Kinetic energy at A [where cos ut — 1, sin ut = 0] = 4mw 2 6 2 
Kinetic energy at B [where cos ut = 0, sin ut = 1] = ^mu 2 a 2 

(6) Method 1. From part (6) of Problem 2.2, 

= J F • dr = \ (—mu 2 r) • dr = — raw 2 I r • dr 

J B B 

d(r • r) = —^mu 2 r 2 
A A 

= \mu 2 a 2 - |mo> 2 6 2 = ^mw 2 (a 2 - 6 2 ) 



Work done 



Method 2. We can assume that at A and B, t = and t = tt/2u respectively. Then: 

Work done = I F • dr 
J A 

Xir/2u 
[— mu 2 (a cos ut i + b sin ut j)] • [—ua sin ut i + ub cos ut j] dt 

s>ir/2(>) 

= I mw 3 (a 2 — 6 2 ) sin wt cos ut dt 



o 

ir/2w 

= %mu 2 (a 2 - b 2 ) sin 2 ut = %mu 2 (a 2 - b 2 ) 

o 

(c) From parts (a) and (6), 

Work done = |ww 2 (a 2 - 6 2 ) = \mu 2 a 2 - \mu 2 b 2 

— kinetic energy at A — kinetic energy at B 

(d) Using Method 2 of part (6) we have, since t goes from to t = 2v/u for a complete circuit 
around the ellipse, 



27T/W 

= 



J-..S7JVCU 
mu z (a 2 — b 2 ) sin ut cos wt dt 
o 

= £mw 2 (a 2 - 6 2 ) sin 2 ut 
Method 1 can also be used to show the same result. 



2.11. Prove that if F is the force acting on a particle and v is the (instantaneous) velocity 
of the particle, then the (instantaneous) power applied to the particle is given by 

cp - f-v 

By definition the work done by a force F in giving a particle a displacement dr is 

dW = F*dr 



Then the (instantaneous) power is given by 
as required. 



* = ^ = '•! = -' 



CHAP. 2] NEWTON'S LAWS OF MOTION. WORK, ENERGY AND MOMENTUM 43 

2.12. Find the (instantaneous) power applied to the particle in Problem 2.1 by the force 
field. 

By Problem 2.1, the velocity and force are given respectively by 

v - (6* 2 + l)i + (12*3 _ 2t)j - 24<k 
F = 60ti + (180* 2 - 10)j - 120k 
Then the power [by Problem 2.11] is given by 

q> = F • v = (60t)(6t 2 + 1) + (180t 2 - 10)(12*3 - 2t) + (120)(24t) 
= 2160t 5 - 120* 3 + 2960£ 

2.13. Find the work done by the force in (a) Problem 2.6, (b) Problem 2.7. 

(a) In the cgs system: v x = |v x | = 1.5 X 10 3 cm/sec, v 2 = |v 2 | = 3.0 X 10 3 cm/sec, m = 10 4 gm. 
Then by Problem 2.8, 

Work done = change in kinetic energy 

cm 2 
= £(10 4 gm)(9.0 X 106 _ 2.25 X 106) _^ 

= 3.38 X lOio V£°*L = 3. 38xlo io/sn_|n N ) (cm) 
sec 2 \ sec z / 

= 3.38 X 10 10 dyne cm = 3.38 X 10 10 ergs 
In the mks system we have similarly: 

Work done = £(10 kg)(900 - 225) ^ 

/ kg m \ 
= 3.38 X 10 3 ( -j^g- ) (m) = 3 - 38 x 10 newton meters 

(6) As in part (a), ». 2 

Work done = £(2000 lb)(88 2 - 2 ) -i^ 



= 7.74 X 106(ft) (^f) = 7.74 X 10« ft pdl 



CONSERVATIVE FORCE FIELDS, POTENTIAL ENERGY, AND 
CONSERVATION OF ENERGY 

2.14. Show that the force field F defined by 

F = (y 2 z 8 - 6xz 2 )i + 2xyz 3 j + {Sxy 2 z 2 - 6x 2 z)k 
is a conservative force field. 
Method 1. The force field F is conservative if and only if curl F = V X F = 0. Now 



V X F = 



i j k 

d/3x d/dy d/dz 

y 2 z z _ q xz z 2xyz s 3xy 2 z 2 - 6x 2 z 



= i \j-{3xy 2 z 2 -6x 2 z) - -j^(2xyzS) 



+ j \^-(y 2 z^-6xz 2 ) - ^(SxyW - §x 2 z) 1 



9a? 



+ k ["^(fcsy* 8 ) - j- (y 2 z* - 6xz 2 )~\ 



Then the force field is conservative. 



44 NEWTON'S LAWS OF MOTION. WORK, ENERGY AND MOMENTUM [CHAP. 2 

Method 2. 

The force field F is conservative if and only if there exists a scalar function or potential 
V(x,y, z) such that F = -grad V = -W Then 

dx dy i dz K 

= (y 2 z 3 - 6xz 2 )i + 2xyz 3 j + (3xy 2 z 2 - 6x 2 z)k 
Hence if F is conservative we must be able to find V such that 

dV/dx = 6xz 2 - y 2 z 3 , dV/dy = -2xyz 3 , BV/dz = 6x 2 z - 3xy 2 z 2 (1) 

Integrate the first equation with respect to x keeping y and z constant. Then 

V = 3x 2 z 2 - xy 2 z 3 + g x {y, z) (2) 

where g t (y, z) is a function of y and z. 

Similarly integrating the second equation with respect to y (keeping x and z constant) and the 
third equation with respect to z (keeping x and y constant), we have 

V = -xy 2 z 3 + g 2 (x, z) (3) 

V = Zx 2 z 2 — xy 2 z 3 + g 3 (x, y) (4) 
Equations (2), (3) and (4) yield a common V if we choose 

ffi(V, *) = c, g 2 (x, z) = 3x 2 z 2 + c, g 3 (x, y) - c (5) 

where c is any arbitrary constant, and it follows that 

V = 3x 2 z 2 — xy 2 z 3 + c 
is the required potential. 

Method 3. 



F-dr = - I (2/2*3 _ Qxz 2 )dx + 2xyz 3 dy + (3xy 2 z 2 - 6x 2 z)dz 

.'o ^ (.x ,y ,z Q ) 

s*(x,y,z) 

= -I d(xy 2 z 3 - 3x 2 z 2 ) = 3x 2 z 2 - xy 2 z 3 + c 



(x ,y ,z ) 

where c = x y 2 z 3 - 3x 2 z^ 



2.15. Prove Theorem 2.2, page 35: If the force acting on a particle is given by F = — vV", 
then the total work done in moving the particle along a curve C from Pi to P% is 

W 



We have 



= f F-rfr = V(Pi) - V(P 2 ) 
= J F-dr = J -VV*dr = 1 



W = \ F-dr = I -VV*dr = I -dV 
Pi Pi J Pi 



= V(P X ) - V(P 2 ) 
Pi 



2.16. Find the work done by the force field F of Problem 2.14 in moving a particle from 
the point A(-2,l,3) to 5(1,-2,-1). 



Work done = f F'dr = f -VV - dr 

J A J A 



J (1,-2,-1) 
-dV = -V(x,y,z) 

(-2,1,3) 

= — 3x 2 z 2 + xy 2 z 3 — c 



(1,-2,-1) 

(-2,1,3) 

(1,-2,-1) 

= 155 

(-2,1,3) 



CHAP. 2] NEWTON'S LAWS OF MOTION. WORK, ENERGY AND MOMENTUM 45 

2.17. (a) Show that the force field of Problem 2.2 is conservative. 
(6) Find the potential energy at points A and B of Fig. 2-3. 

(c) Find the work done by the force in moving the particle from A to B and compare 
with Problem 2.10(6). 

(d) Find the total energy of the particle and show that it is constant, i.e. demonstrate 
the principle of conservation of energy. 

(a) From Problem 2(6), F = —ma 2 r = -ma 2 {xi + yj). Then 

i J k 



V X F = 



d/dx d/dy d/dz 

—ma 2 x —mu 2 y 



+ k \h { ~ mu2y) ~ ^ { ~ ma2x) \ 



= 

Hence the field is conservative. 

(6) Since the field is conservative there exists a potential V such that 

W- dV - dV u 

Then dV/dx = ma 2 x, dV/dy = mw 2 y, dV/dz = 

from which, omitting the constant, we have 

V = \ma 2 x 2 + %ma 2 y 2 = $ma 2 (x 2 + y 2 ) = \mJh 2 
which is the required potential. 

(c) Potential at point A of Fig. 2-3 [where r = a] = $ma 2 a 2 . 
Potential at point B of Fig. 2-3 [where r = b] = £mu 2 6 2 . Then 

Work done from A to B = Potential at A — Potential at B 

= \ma 2 a 2 - £m<o 2 6 2 = $ma 2 (a 2 - b 2 ) 
agreeing with Problem 2.10(6). 

(d) By Problems 2.10(a) and part (6), 

Kinetic energy at any point = T = %mv 2 = |mf 2 

= |ra(<o 2 a 2 sin 2 at + a 2 b 2 cos 2 at) 

Potential energy at any point = V = ^m« 2 r 2 

= %ma 2 (a 2 cos 2 at + b 2 sin 2 at) 

Thus at any point we have on adding and using sin 2 at + cos 2 at = 1, 

T+V = $ma 2 (a 2 + b 2 ) 
which is a constant. 

IMPULSE, TORQUE, ANGULAR MOMENTUM, AND 
CONSERVATION OF MOMENTUM 

2.18. Prove Theorem 2.6, page 36: The impulse of a force is equal to the change in 
momentum. 

By definition of impulse [see (19), page 36] and Newton's second law, we have 

t2 



C* ¥dt = J 2 -^( m v) dt = J d ^ 



(mv) 



= mv 2 — mxi 



46 NEWTON'S LAWS OF MOTION. WORK, ENERGY AND MOMENTUM [CHAP. 2 

2.19. A mass of 5000 kg moves on a straight line from a speed of 540 km/hr to 720 km/hr 
in 2 minutes. What is the impulse developed in this time? 

Method 1. 

Assume that the mass travels in the direction of the positive x axis. In the mks system, 
_ _. n . km _ 540i X 1000 m 1Kvin2 . m 

V, = 5401-:; = = 1.5 X 10 2 1 

hr 3600 sec sec 

„™. km 720i X 1000 m nn „««. m 

v 2 = 720i=^ = = 2.0Xl0 2 i — 

hr 3600 sec sec 

Then from Problem 2.18, 

Impulse = m(v 2 — v x ) = (5000 kg)(0.5 X 10 2 i m/sec) 

= 2.5 X 10 5 i kg m/sec = 2.5 X 10 5 i newton sec 

since 1 newton = 1 kg m/sec 2 or 1 newton sec = 1 kg m/sec. 

Thus the impulse has magnitude 2.5 X 10 5 newton sec in the positive x direction. 

Method 2. 

Using the cgs system, v x = 540i km/hr = 1.5 X 10 4 i cm/sec and v 2 = 720i km/hr = 
2.0 X 10 4 i cm/sec. Then 

Impulse = w(v 2 - v t ) = (5000 X 10 3 gm)(0.5 X 10 4 i cm/sec) 

= 2.50 X 10 10 igm cm/sec = 2.50 X 10 10 i dyne sec 

since 1 dyne = 1 gm cm/sec 2 or 1 dyne sec = 1 gm cm/sec 

Note that in finding the impulse we did not have to use the time 2 minutes as given in the 
statement of the problem. 



2.20. Prove Theorem 2.7, page 36: The moment of force or torque about the origin O of 
a coordinate system is equal to the time rate of change of angular momentum. 

The moment of force or torque about the origin O is 

A = rXF = rX-j- (mv) 

The angular momentum or moment of momentum about O is 

B = m(r X v) = r X (mv) 

Now we have 42L = ^(rXmv) = J^ X (mv) + rx|(mv) 

= v X (mv) + r X -(mv) = + rXF = A 
at 

which gives the required result. 

2.21. Determine (a) the torque and (6) the angular momentum about the origin for the 
particle of Problem 2.4 at any time t. 

(a) Torque A = rXF 

= [(* 4 + 6* + 3)i + (3*3 - 4* 2 + 15* - 1) j + (4 - * 3 - 8*)k] X [24* 2 i + (36* - 16)j - 12*k] 

i J k 

t 4 + 6* + 3 3* 3 - 4* 2 + 15* - 1 4 - t 3 - 8* 

24t 2 36* - 16 -12* 

= (32*3 + 108*2 _ 260* + 64)i - (12* 5 + 192* 3 - 168* 2 - 36*) j 

- (36* 5 - 80* 4 + 360*3 _ 240* 2 - 12* + 48)k 



CHAP. 2] NEWTON'S LAWS OF MOTION. WORK, ENERGY AND MOMENTUM 



47 



(b) Angular momentum SI = r X (mv) = m(r X v) 

= 2[(t* + 6* + 3)i + (3*3 - 4* 2 + 15* - l)j + (4 - *3 - 8*)k] 
X [(4*3 + 6)i + (9* 2 - 8* + 15)j - (3*2 + 8)k] 

i J k 

= 2 * 4 + 6* + 3 3* 3 - 4*2 + 15* - 1 4 - * 3 - 8* 
4*3 + 6 9* 2 - 8* + 15 -3* 2 - 8 

= (8*4 + 36*3 _ 130*2 + 64t _ I04)i - (2*« + 48** - 56* 3 - 18* 2 - 96)j 
- (6*« - 16*5 + 90*4 _ 80*3 _ 6*2 + 48* - 102)k 

Note that the torque is the derivative with respect to * of the angular momentum, illustrating 
the theorem of Problem 2.20. 

2.22. A particle moves in a force field given by F = r 2 r where r is the position vector of 
the particle. Prove that the angular momentum of the particle is conserved. 
The torque acting on the particle is 

A = rXF = rX (r 2 r) = r 2 (r X r) = 

Then by Theorem 2.9, page 37, the angular momentum is constant, i.e. the angular momentum is 
conserved. 



NON-CONSERVATIVE FORCES 

2.23. Show that the force field given by F = x 2 yzi - xyz^k is non-conservative. 

We have i j k 

V X F = d/dx d/dy BlBz 
x 2 yz —xyz 2 

Then since V X F # 0, the field is non-conservative. 



= -xz*\ + (x 2 y + yz 2 )} - x 2 zk 



STATICS OF A PARTICLE 

2.24. A particle P is acted upon by the forces Fi, F 2 , F 3 , F 4 , F 8 and F 6 shown in Fig. 2-6. 
Represent geometrically the force needed to prevent P from moving. 




Fig. 2-6 




Fig. 2-7 



The resultant R of the forces F v F 2 , F 3 , F 4 , F 5 and F 6 can be found by vector addition as 
indicated in Fig. 2-7. We have R = F 1 + F 2 + F 3 -|-F4 + F5 + F 6 . The force needed to prevent 
P from moving is — R which is a vector equal in magnitude to R but opposite in direction and 
sometimes called the equilibrant. 



48 



NEWTON'S LAWS OF MOTION. WORK, ENERGY AND MOMENTUM [CHAP. 2 



2.25. A particle is acted upon by the forces Fi = 5i - lOj + 15k, F 2 = lOi + 25j - 20k and 
F 3 = 15i-20j + 10k. Find the force needed to keep the particle in equilibrium. 

The resultant of the forces is 

R = F x + F 2 + F 3 = (5i - lOj + 15k) + (lOi + 25j - 20k) + (151 - 20j + 10k) 
= 30i - 5j + 5k 
Then the force needed to keep the particle in equilibrium is — R = — 30i + 5 j — 5k. 

2.26. The coplanar forces as indicated in Fig. 2-8 act on a particle P. Find the resultant 
of these forces (a) analytically and (b) graphically. What force is needed to keep 
the particle in equilibrium? 



Unit = 20 lb 





V 




''/ 


30 °/\ 


V 


•3/ 




.#/- 




///30° 




iV 






Fig. 2-8 Fig. 2-9 

(a) Analytically. From Fig. 2-8 we have, 

F x = 160(cos45° i + sin 45° j), F 2 = 100(- cos 30° i + sin 30° j), 
F 3 = 120(- cos 60° i - sin 60° j) 
Then the resultant R is 
R = F t + F 2 + F 3 

= (160 cos 45° - 100 cos 30° - 120 cos 60°)i + (160 sin 45° + 100 sin 30° - 120 sin 60°)j 
= -33.46i + 59.21J 

Writing R = R cos a i + R sin a j where a is the angle with the positive x axis measured 
counterclockwise, we see that 

R cos a - -33.46, R sin a = 59.21 



Thus the magnitude of R is R = V(-33.46) 2 + (59.21) 2 = 68.0 lb, and the direction a with 
the positive x axis is given by tan a = 59.21/(-33.46) = -1.770 or a = 119° 28'. 

(6) Graphically. Choosing a unit of 20 lb as shown in Fig. 2-9, we find that the resultant has 
magnitude of about 68 lb and direction making an angle of about 61° with the negative 
x axis [using a protractor] so that the angle with the positive x axis is about 119°. 

A force — R, i.e. opposite in direction to R but with equal magnitude, is needed to keep P 
in equilibrium. 



STABILITY OF EQUILIBRIUM 

2.27. A particle moves along the x axis in a force field having potential V = fax 2 , k > 0. 
(a) Determine the points of equilibrium and (5) investigate the stability. 

(a) Equilibrium points occur where VV = or in this case 

dV/dx = kx = or x = 

Thus there is only one equilibrium point, at x = 0. 



CHAP. 2] NEWTON'S LAWS OF MOTION. WORK, ENERGY AND MOMENTUM 



49 



(6) Method 1. 

Since d 2 V/dx 2 = k > 0, it follows that at x = 0, V is a minimum. Thus by Theorem 2.10, 

page 38, * = is a point of stability. This is also seen from Problem 2.36 where it is shown 

that the particle oscillates about x = 0. 

V(x) 



Method 2. 



dV. 



= — kxi. Then 



We have F = -VV = — ^-i = 
ax 

when x > the particle undergoes a force to 
the left, and when x < the particle under- 
goes a force to the right. Thus x = is a 
point of stability. 

Method 3. 

The fact that x = is a minimum point can 
be seen from a graph of V(x) vs x [Fig. 2-10]. 




Fig. 2-10 



MISCELLANEOUS PROBLEMS 

2.28. Show how Newton's laws can be used to develop definitions of force and mass. 

Let us first consider some given particle P, assuming for the present that its mass ra P is not 
defined but is simply some constant scalar quantity associated with P. Axiom 1 states that if P 
moves with constant velocity (which may be zero) then the force acting on it is zero. Axiom 2 
states that if the velocity is not constant then there is a force acting on P given by m P a P where 
a P is the acceleration of P. Thus force is defined by axioms 1 and 2 [although axiom 1 is unnecessary 
since it can in fact be deduced from axiom 2 by letting F = 0]. It should be noted that force is a 
vector and thus has all the properties of vectors, in particular the parallelogram law for vector 
addition. 

To define the mass m P of particle P, let us now allow it to interact with some particular 
particle which we shall consider to be a standard particle and which we take to have unit mass. 
If a P and a s are the accelerations of particle P and the standard particle respectively, it follows 
from axioms 2 and 3 that m P a P = — a s . Thus the mass m P can be defined as — a s /a P . 



2.29. Find the work done in moving a particle once around a circle C in the xy plane, if 
the circle has center at the origin and radius 3 and if the force field is given by 

F = (2x-y + z)i + (x + y-z 2 )j + (Sx - 2y + 4z)k 

In the plane z = 0, F = (2x - y)\ + (x + y)j + (3x - 2y)k and rfr = dx i + dy j so that the 
work done is 

f F'dr - f [(2x-y)i + (x + y)j + (Sx-2y)k]'[dxi + dyj] 
J c J c 

= I (2x — y) dx + (x + y) dy 
J c 

Choose the parametric equations of the circle as x = 3 cos t, 
y = 3 sin t where t varies from to 2v [see Fig. 2-11]. Then the 
line integral equals 

,2tt 



I [2(3 cos t) - 3 sin t] [-3 sin t] dt + [3 cos t + 3 sin t] [3 cos t] dt 

X27T Q 

(9-9 sin t cos t) dt = 9t - - sin 2 1 



= lSir 



In traversing C we have chosen the counterclockwise direction indi- 
cated in Fig. 2-11. We call this the positive direction, or say that C 
has been traversed in the positive sense. If C were traversed in the 
clockwise (negative) direction the value of the integral would be — 18tt. 




r = xi + yj 

= 3 cos t i + 3 sin t j 

Fig.2-ll 



50 NEWTON'S LAWS OF MOTION. WORK, ENERGY AND MOMENTUM [CHAP. 2 

2.30. (a) If F = -vV r , where V is single-valued and has continuous partial derivatives, 
show that the work done in moving a particle from one point Pi s (xi, yi, zi) in 
this field to another point P 2 = (x 2 , 2/2, z 2 ) is independent of the path joining the 
two points. 

(b) Conversely, if J F'dr is independent of the path C joining any two points, 
show that there exists a function V such that F=-vV. 



ft pP 2 

(a) Work done = I F«dr = — I VV»dr 
J *x J P! 

" -J Pi (to 1 + aP + -^kj-idxi + dyj + dzk) 

J Pl d % ty dz 

= -f 2<iV = V(P 1 )-V(P 2 ) = V(x x ,y x ,z x ) - V(x 2 ,y 2 ,z 2 ) 



Then the integral depends only on points P x and P 2 and not on the path joining them. 
This is true of course only if V(x,y,z) is single-valued at all points P x and P 2 . 

(6) Let F = F x \ + F 2 j + F 3 k. By hypothesis, I F • dx is independent of the path C joining any 
two points, which we take as (x x ,y x , z x ) and (x,y,z) respectively. Then 

J % (x,y,z) p(.x,y,z) 

F • dt = - I (F x dx + F 2 dy + F s dz) 

(.x i ,y 1 ,z 1 '> J <.x v y lt zi> 

is independent of the path joining (x x , y x , z x ) and (x,y,z). Thus 

V(x, y,z) = - I [F x (x, y, z) dx + F 2 (x, y, z) dy + F 3 (x, y, z) dz] 

where C is a path joining (x x ,y x ,z x ) and (x,y,z). Let us choose as a particular path the 
straight line segments from (x it y v z x ) to (x, y x , z x ) to (x, y, z x ) to (x, y, z) and call V(x, y, z) the 
work done along this particular path. Then 

XX pV pz 

F x (x,y x ,z x )dx - J F 2 (x,y,z x )dy - J F 3 (x,y,z)dz 
"i *i ^1 

It follows that 
— = -F 3 {x,y,z) 

dV C dF 3 C z 9F 2 

7- = -F2(x,y,z x ) - I — — (x,y,z)dz = -F 2 {x,y,z x ) - I — -(x,y,z)dz 

dy ^ Xl dy *s Zl dz 

z 

= -F 2 (x,y,z x ) - F 2 (x,y,z) 

*i 

= -F 2 (x,y,z x ) - F 2 (x,y,z) + F 2 (x,y,z x ) = -F 2 (x,y,z) 



§V_ 
dx 



C v SF 2 f 8F a 

= -f x (x, y x , z x ) - j -^-(*> v> *i) d y - j -fate* y> *) dz 

J'* dF x C* dF x 

-r-{x,y,z x )dy - I — (x,y,z) dz 
j/j <>y ^ zx °Z 

y \z 

= -F x (x,y x ,z x ) - F x (x,y,z x ) - F x (x,y,z)\ 

v\ \*i 

= -F x (x,y x ,z x ) - F x (x,y,z x ) + F(x,y x ,z x ) - F x (x,y,z) + F(x,y,z x ) = -F x (x,y,z) 



CHAP. 2] NEWTON'S LAWS OF MOTION. WORK, ENERGY AND MOMENTUM 



51 



Then 



F = r.l+J-.J+F.k = -fi-fi-fk = -W 



3»" dy' dz 
Thus a necessary and sufficient condition that a field F be conservative is that curl F = V X F = 0. 



2.31. (a) Show that F = (2xy + z*)i + x 2 j + Sxz 2 k is a conservative force field, (b) Find 
the potential, (c) Find the work done in moving an object in this field from (1, -2, 1) 

to (3,1,4). 



(a) A necessary and sufficient condition that a force will be conservative is that curl F = V X F — 0. 

= 0. Thus F is a conservative force field. 



Now V X F = 



i J k 

d/dx d/dy d/dz 

2xy + z* x 2 Zxz 2 



(6) As in Problem 2.14, Methods 2 or 3, we find V = ~(x 2 y + xz 3 ). 

(3,1,4) 



(c) Work done = — (x 2 y + xz 3 ) 



= -202. 



(1,-2,1) 



F'dr is independent of the path joining any two points Pi and P 2 

Pi r 

in a given region, then i F • dt = f or all closed paths in the region and conversely. 

Let P 1 AP 2 BP 1 [see Fig. 2-12] be a closed curve. Then ^^^ 

£ F*dr = J F • dr = J F • dr + J F • dr ^^^^^^ \p 

J PiAPtBP! PiAP 2 P2BP1 S^ \ 2 

= J F-dr- J F-dr = 



PiAP 2 PiBPa 

since the integral from P x to P 2 along a path through A is 
the same as that along a path through B, by hypothesis. 



L^ 



Conversely if <i> F*dr = 0, then 

J F'dr = J F'dr + J F'dr = J F»dr - J 

P^PiBP! PiAP 2 

, $ F-dr = / F 



A 
Fig. 2-12 



F • dr = 



P 2 BP t 



P1AP2 



P 1 BP 2 



so that, I F»dr = I F«dr 

PiAP 2 PiBP 2 



2.33. (a) Show that a necessary and sufficient condition that Fi dx + F 2 dy + Fs dz be an 
exact differential is that VxF = where F = Fii + F 2 j+F 3 k. 

(b) Show that (y 2 z? cos x - 4x s z) dx + 2z s y sin x dy + (Sy 2 z 2 sin x - x 4 ) dz is an exact 
differential of a function <f> and find <f>. 

(a) Suppose F x dx + F 2 dy + F 3 dz = d<f> = -^-dx + -^-dy + -^-dz, an exact differential. Then 
since x, y and z are independent variables, 



52 NEWTON'S LAWS OF MOTION. WORK, ENERGY AND MOMENTUM [CHAP. 2 



30 jt, _ d<f> „ _ 30 

dx dy dz 



and so F = F x i + F 2 j + F s k = |*i + |* j + |^k = V*. Thus VXF = VXV^ = 0. 

dx dy dz ^ 

Conversely if V X F = 0, then F = V> and so F'dr = V#*dr = d<f>, i.e. Fj da; + 
F 2 dy + F 3 dz = d<p, an exact differential. 

(b) F = (y 2 z s cos a; — Ax z z) i + 2z 3 j/ sin x j + (32/ 2 « 2 sin x — x*) k and V X F is computed to be 
zero, so that by part (a) the required result follows. 

2.34. Referring to Problem 2.4 find (a) the kinetic energy of the particle at t - 1 and 
£ = 2, (6) the work done by the field in moving the particle from the point where 
t = 1 to the point where t = 2, (c) the momentum of the particle at t = 1 and t = 2 
and (d) the impulse in moving the particle from t = 1 to £ = 2. 

(a) From part (a) of Problem 2.4, 

v = (4*3 + 6)i + (9i 2 - 8t + 15)j - (St 2 + 8)k 
Then the velocities at t = 1 and t = 2 are 

v x = lOi + 16j - Ilk, v 2 = 38i + 35j - 20k 
and the kinetic energies at t = 1 and t = 2 are 

T x = £ mv ? = £(2)[(10)2 + (16)2 + (-11)2] = 477, T 2 = £mv 2 = 3069 

(6) Work done = f F • dr 



-s. 

-X 



[24* 2 i + (36t - 16)j - 12tk] • [(4« 3 + 6)i + (9* 2 - 8t + 15)j - (3t 2 + 8)k]dt 

2 

[(24t 2 K4*3 + 6) 4- (36£ - 16)(9t 2 -8t + 15) + (12t)(3« 2 + g)]dt = 2592 



Note that by part (a) this is the same as the difference or change in kinetic energies 
3069-477 = 2592, illustrating Theorem 2.1, page 35, that Work done = change in kinetic 
energy. 

(c) By part (a) the momentum at any time t is 

p = mv = 2v = (8« 3 + 12)i + (18t 2 - 16* + 30)j - (6t 2 + 16)k 

Then the momenta at t = 1 and t = 2 are 

Pi = 20i + 32j - 22k, p 2 = 76i + 70j - 40k 

(d) Impulse = ) F dt 

J t=i 

.2 

[24t*i + (36t - 16) j - 12tk]dt = 56i + 38 j - 18k 
't=i 

Note that by part (6) this is the same as the difference or change in momentum, i.e. 
P2 ~ Pi = ( 76i + 7 °J ~ 40k ) - ( 20i + 32 J ~~ 22k > = 56i + 38 ' ~ 18k ' illu strating Theorem 2.6, 
page 36, that Impulse = change in momentum. 



=/,: 



2.35. A particle of mass m moves along the x axis under the influence of a conservative 
force field having potential V(x). If the particle is located at positions x x and x 2 at 
respective times U and U, prove that if E is the total energy, 

♦ * - [™ C H dx 

* 2 " *» " \ 2 X t y/E - V(x) 



CHAP. 2] NEWTON'S LAWS OF MOTION. WORK, ENERGY AND MOMENTUM 53 

By the conservation of energy, 

Kinetic energy + Potential energy = E 
$m{dx/dt) 2 + V{x) = E 

Then {dx/dt) 2 = (2/m){E - V(x)} (1) 

from which we obtain on considering the positive square root, 

dt = yfm/2(dx/\/E-V{x)) 
Hence by integration, 



J ti dt ~ **-** - \f f Xi Ve^vW) 



2.36. (a) If the particle of Problem 2.35 has potential V = fax 2 and starts from rest at 
x = a, prove that x = acosy^Jmt and (b) describe the motion. 

(a) From (i) of Problem 2.35, (dx/dt) 2 = (2/m)(E - fax 2 ). Since dx/dt = where x = a, we 
find E - faa 2 so that 

(dx/dt) 2 = (K/m)(a 2 -x 2 ) or dx/y/a 2 - x 2 = ±y/Umdt 

Integration yields sin -1 (x/a) = ± y/ic/m t + c x . Since x = a at t = 0, c l = v/2. Then 
sin _1 (*/a) = ± sjic/m t + jt/2 or * = a sin (jt/2 ± y/ic/m t) = a co^yfic/mt 

(6) The particle oscillates back and forth along the x axis from x = a to x — —a. The time for 
one complete vibration or oscillation from x — a back to x = a again is called the period of the 
oscillation and is given by P = 2v VWk. 



2.37. A particle of mass 3 units moves in the xy plane under the influence of a force field 
having potential V = l2x{Zy - 4x). The particle starts at time t = from rest at 
the point with position vector lOi - lOj. (a) Set up the differential equations and 
conditions describing the motion. (6) Solve the equations in (a), (c) Find the 
position at any time, (d) Find the velocity at any time. 

(a) Since V = 12x(3y - 4*) = S6xy - 48a 2 , the force field is 

F = ~ Vy = -f^-^-f 1 ' = (-362/ + 96*)i-36*j 



Then by Newton's second law, 



d 2 * 
3 ;/72 ~ (~36y + 96x)i - 36;rj 



dt 2 
or in component form, using r = xi + yj, 

dtx/dt 2 = -12y + 32a, cPy/dt 2 = -12* (1) 

where x = 10, x = 0, y = -10, y = at t = (2) 

using the fact that the particle starts at r = lOi — lOj with velocity v = f = 0. 

(6) From the second equation of (i), * = —^d 2 yfdt 2 . Substitution into the first equation of (1) 
yields 

d*y/dt* - 32 d 2 y/dt 2 - 144y = (3) 

If a is constant then y = e«* is a solution of ($) provided that 

«4 - 32a 2 - 144 = 0, i.e. (a 2 + 4)(a 2 - 36) = or a = ±2i, a = ±6 



54 NEWTON'S LAWS OF MOTION. WORK, ENERGY AND MOMENTUM [CHAP. 2 

Thus solutions are e 2it , e~ 2it , e 6t , e~ 6t or cos2t, sin2t, e 6t , e -6t [in terms of real functions] 
and the general solution is 

y = c t cos2t + c 2 sin2* + c 3 e 6t + c 4 e~ 6t (4) 

Thus from a; = ~^cPy/dt 2 we find, using (.4), 

a? = ^ Cos 2t + \c 2 sin 2* - 3c 3 e« - 3c 4 e~ 6t (5) 

Using the conditions (2) in (4) and (5), we obtain 

£<?! - 3c 3 - 3c 4 = 10, §c 2 - 18c 3 + 18c 4 = 0, 
c i + c 3 + c 4 = — 10, 2c 2 + 6c 3 — 6c 4 = 
Solving simultaneously, c x = —6, c 2 = 0, c 3 = —2, c 4 = —2 so that 

a? = -6 cos2t- 2e 6t - 2e~ 6 *, y = -2 cos 2t + 6e 6t + 6e-« 

(c) The position at any time is 

r = xi + yj = (-6 cos2t - 2e 6t - 2e~ flt )i + (-2 cos 2* + 6e 6t + 6e~ 6t )j 

(d) The velocity at any time is 

v = r = xi + yj = (12 sin 2* - 12e«' + 12e" 6t )i + (4 sin 2* + 36e 6 ' - 36e~ 6t )j 

In terms of the hyperbolic functions 

sinh at = £(««*- «-«*)» cosh at = ^(e at -\- e~ at ) 

we can also write 

r = (-6 cos 2t - 4 cosh 6t)i + (-2 cos 2t + 12 cosh 6t)j 

v = r = (12 sin 2t - 24 sinh 6t)i + (4 sin It + 72 sinh 6«)j 

2.38. Prove that in polar coordinates (r, 0), 

Let VV = Gri + H*! (i) 

where G and H are to be determined. Since dr = dx i + dy j we have on using a? = r cos , 
y = r sine and Problem 1.47(6), page 25, 

dr = (cos * dr — r sin *)(cos $ r x — sin 9^ + (sin tf dr + r cos d*)(sin e r t + cos # x ) 
or dr = dr r x + r dff # x (2) 

Now VV'dr = dV = -^dr + -^de 

Using (i) and (2) this becomes 

(Gr x + H9 1 )-(drr 1 + rd69i) = Gdr + Hrde = jp dr + 'gf de 

so that G ~ Ifr' H ~ rle 

Then (1) becomes VV = ^T r i + ^ a7 # i 

2.39. According to the theory of relativity, the mass m of a particle is given by 

— m ° mp 

W ~ VI - vVc 2 VI -)8 2 
where v is the speed, m the rest mass, c the speed of light and p = v/c. 



CHAP. 2] NEWTON'S LAWS OF MOTION. WORK, ENERGY AND MOMENTUM 55 

(a) Show that the time rate of doing work is given by 

moc 2 ^!-/? 2 )-" 2 

(b) Deduce from (a) that the kinetic energy is 

T = (m-mo)c 2 = m c 2 {(l-/3 2 )- 1/2 -l} 

(c) If v is much less than c, show that T = %mv 2 approximately. 
(a) By Newton's second law, , , / m v 



F = -jT(mv) = 






dt dt \y/± 

Then if W is the work done, 

dW _ d / "W \ d f P \ 2 d / 1 \ 

as proved by direct differentiation. 

(6) Since Work done = change in kinetic energy, we have 

Time rate of doing work = time rate of change in kinetic energy 

u _* / x dW dT 2 d / 1 \ 

or by part (a), ~dT = Hi = m « c *dt (^f^p) 

WoC 2 

Integrating, T = + c x 

To determine Cj note that, by definition, T = when v = or /? = 0, so that Cj = — m « 
Hence we have, as required, 

T = ——z^^ — m c 2 = (m — mo)c 2 
(c) For /? < 1 we have by the binomial theorem, 

1 = (I-/?*)-"* = 1 + 1^2 + 1^1^ + 1^111^+... 



r i v 2 i i 

Then T = m c 2 1 + - -^ + • • • — m c 2 = -^mv 2 approximately 



$ Supplementary Problems 

NEWTON'S LAWS 

2.40. A particle of mass 2 units moves along the space curve defined by r = (4t 2 — t 3 )i — 5tj + (t* — 2)k. 
Find (a) the momentum and (6) the force acting on it at t = 1. 

Ans. (a) 10i-10j + 8k, (6) 4i + 24k 

2.41. A particle moving in a force field F has its momentum given at any time t by 

p = 3e _t i — 2 cos t j — 3 sin t k 
Find F. Ans. — 3e~*i + 2 sin £ j — 3 cos t k 

2.42. Under the influence of a force field a particle of mass m moves along the ellipse 

r = a cos at i + b sin ut j 
If p is the momentum, prove that (o)rXp = mabak, (6) r • p = £m(& 2 — a 2 ) sin 2at. 



56 NEWTON'S LAWS OF MOTION. WORK, ENERGY AND MOMENTUM [CHAP. 2 

2.43. If F is the force acting on the particle of Problem 2.42, prove that r X F = 0. Explain what this 
means physically. 

2.44. A force of 100 dynes in the direction of the positive x axis acts on a particle of mass 2 gm for 
10 minutes. What velocity does the particle acquire assuming that it starts from rest? 

Arts. 3 X 10 4 cm/sec 

2.45. Work Problem 2.44 if the force is 20 newtons and the mass is 10 kg. Ans. 1200 m/sec 

2.46. (a) Find the constant force needed to accelerate a mass of 40 kg from the velocity 4i — 5j + 3k 
m/sec to 8i + 3j — 5k m/sec in 20 seconds. (6) What is the magnitude of the force in (a)? 
Ans. (a) 8i + 16j - 16k newtons or (8i + 16j - 16k) X 10 5 dynes 

(6) 24 newtons or 24 X 10 5 dynes 

2.47. An elevator moves from the top floor of a tall building to the ground floor without stopping. 
(a) Explain why a blindfolded person in the elevator may believe that the elevator is not moving 
at all. (b) Can the person tell when the motion begins or stops? Explain. 

2.48. A particle of unit mass moves in a force field given in terms of time t by 

F = (6«-8)i - 60t3j + (20t3 + 36«2)k 

Its initial position and velocity are given respectively by r = 2i — 3k and v = 5i + 4j. Find the 
(a) position and (b) velocity of the particle at t = 2. 
Ans. (a) 4i - 88j + 77k, (6) i - 236j + 176k 

2.49. The force acting on a particle of mass m is given in terms of time t by 

F = a cos ut i + b sin at j 
If the particle is initially at rest at the origin, find its (a) position and (6) velocity at any later time. 

Am. (a) -— ^-(1 — cos at) i H ^(at — sin at) j, (6) — sin at i H (1 — cos at) j 

ma 2 ma 2 ma ma 

WORK, POWER AND KINETIC ENERGY 

2.50. A particle is moved by a force F = 20i — 30j + 15k along a straight line from point A to point B 
with position vectors 2i + 7j — 3k and 5i — 3 j — 6k respectively. Find the work done. 

Ans. 315 

2.51. Find the kinetic energy of a particle of mass 20 moving with velocity 3i — 5j + 4k. Ans. 500 

2.52. Due to a force field F, a particle of mass 4 moves along the space curve r = (3t 2 — 2t)\ + t 3 j — t 4 k. 
Find the work done by the field in moving the particle from the point where t = 1 to the point 
where t = 2. Ans. 2454 

2.53. At one particular instant of time a particle of mass 10 is traveling along a space curve with velocity 
given by 4i + 16k. At a later instant of time its velocity is 8i — 20j. Find the work done on the 
particle between the two instants of time. Ans. 192 

2.54. Verify Theorem 2.1, page 35 for the particle of Problem 2.52. 

2.55. A particle of mass m moves under the influence of the force field given by F = a(sin at i + cos at j). 
If the particle is initially at rest at the origin, prove that the work done on the particle up to time t 
is given by (a 2 /m<o 2 )(l — cos at). 

2.56. Prove that the instantaneous power applied to the particle in Problem 2.55 is (a 2 /mw) sin ut. 

2.57. A particle moves with velocity 5i-3j + 6k under the influence of a constant force F = 20i + 
10j + 15k. What is the instantaneous power applied to the particle? Ans. 160 



CHAP. 2] NEWTON'S LAWS OF MOTION. WORK, ENERGY AND MOMENTUM 57 

CONSERVATIVE FORCE FIELDS, POTENTIAL ENERGY AND 
CONSERVATION OF ENERGY 

2.58. (a) Prove that the force field F = (y 2 — 2xyz s )i + (3 + 2xy — x 2 z 3 )j + (6z 3 - 3x 2 yz 2 )k is conservative. 
(6) Find the potential V associated with the force field in (a). 

Ans. (b) xy 2 — x 2 yz 3 + 3y + § z 4 

2.59. A particle moves in the force field of Problem 2.58 from the point (2, — 1, 2) to (—1, 3, —2). Find the 
work done. Ans. 55 

2.60. (a) Find constants a, b, c so that the force field defined by 

F = (x + 2y + az)i + (bx -3y- z)j + (4x + cy + 2z)k 
is conservative. 

(6) What is the potential associated with the force field in (a)? 

Ans. (a) a = 4, 6 = 2, c = -1 (6) V = -\x 2 + f y 2 - z 2 - 2xy - 4xz + yz 

2.61. Find the work done in moving a particle from the point (1,-1,2) to (2,3,-1) in a force field with 
potential V = x 3 — y 3 + 2xy — y 2 + 4x. Ans. 15 

2.62. Determine whether the force field F = (x 2 y — z 3 )i + (3xyz + xz 2 )j + (2x 2 yz + j/z 4 )k is conservative. 
Ans. Not conservative 

2.63. Find the work done in moving a particle in the force field F = 3x 2 i + (2xz — y)j + zk along 
(a) the straight line from (0, 0, 0) to (2, 1, 3), (6) the space curve x = 2t 2 , y = t, z = 4t 2 —t from t = 
to t = l. Is the work independent of the path? Explain. Ans. (a) 16, (6) 14.2 

2.64. (a) Evaluate d> F • dt where F = (x — 3y)i + (y — 2#)j and C is the closed curve in the xy plane 

J c 
x — 2 cos t, y = 3 sin t from t = to t = 2v. (b) Give a physical interpretation to the result in (a). 
Ans. (a) 6jt if C is traversed in the positive (counterclockwise) direction. 

2.65. (a) Show that the force field F = — kt^t is conservative. 

(6) Write the potential energy of a particle moving in the force field of (a). 

(c) If a particle at mass m moves with velocity v = dr/dt in this field, show that if E is the constant 
total energy then ^m(dr/dt) z + ^kt 5 = E. What important physical principle does this illustrate? 

2.66. A particle of mass 4 moves in the force field defined by F = — 200r/r3. (a) Show that the field is 
conservative and find the potential energy. (6) If a particle starts at r = 1 with speed 20, what will 
be its speed at r = 2? Ans. (a) V - 200/r, (6) 15^ 

IMPULSE, TORQUE AND ANGULAR MOMENTUM. 
CONSERVATION OF MOMENTUM 

2.67. A particle of unit mass moves in a force field given by F = (3t 2 - te)i + (12t - 6)j + (6* - 12^ 
where t is the time, (a) Find the change in momentum of the particle from time t — 1 to t = 2. 
(6) If the velocity at * = 1 is 4i — 5j + 10k, what is the velocity at * = 2? 

Ans. (a) i + 12j - 19k, (6) 5i + 7j - 9k 

2.68. A particle of mass m moves along a space curve defined by r = a cos at i + 6 sin at j. Find 
(a) the torque and (b) the angular momentum about the origin. Ans. (a) 0, (6) 2mabak 

2.69. A particle moves in a force field given by F = <f>(r) r. Prove that the angular momentum of the 
particle about the origin is constant. 

2.70 Find (a) the torque and (6) the angular momentum about the origin at the time t = 2 for the 
particle of Problem 2.67, assuming that at t = it is located at the origin. 
Ans. (a) - (36i + 128j + 60k), (6) - 44i + 52j + 16k 

2.71. Find the impulse developed by a force given by F = 4£i + (6t 2 — 2)j + 12k from t = to t = 2. 
Ans. 8i + 12j + 24tk 



58 



NEWTON'S LAWS OF MOTION. WORK, ENERGY AND MOMENTUM [CHAP. 2 



2.72. What is the magnitude of the impulse developed by a mass of 200 gm which changes its velocity from 
5i — 3j + 7k m/sec to 2i + 3j + k m/sec? Ans. 1.8 X 10 5 dyne sec or 1.8 newton sec 



STATICS OF A PARTICLE 

2.73. A particle is acted upon by the forces F x = 2i + aj - 3k, F 2 = 5i + cj + 6k, F 3 = 6i - 5j + 7k, 
F 4 = ci — 6j + ak. Find the values of the constants a, b, c in order that the particle will be in 
equilibrium. Ans. a — 1, 6 = 11, c = 4 

2.74. Find (a) graphically and (6) analytically the result- 
ant force acting on the mass m of Fig. 2-13 where 
all forces are in a plane. 

Ans. (6) 19.5 dynes in a direction making an angle 
85°22' with the negative x axis 

2.75. The potential of a particle moving in the xy plane is 
given by V = 2x 2 — 5xy + Sy 2 + 6x-7y. (a) Prove 
that there will be one and only one point at which a 
particle will remain in equilibrium and (6) find the 
coordinates of this point. Ans. (6) (1, 2) 

2.76. Prove that a particle which moves in a force field 
of potential 

V - x 2 + Ay 2 + z 2 - Axy - 4yz + 2xz - 4a; + Sy - 4z 
can remain in equilibrium at infinitely many points 
and locate these points. 
Ans. All points on the plane x — 2y-Vz — 2 




Fig. 2-13 



STABILITY OF EQUILIBRIUM 

2.77. A particle moves on the x axis in a force field having potential V = x 2 (6 — x). 
(a) Find the points at equilibrium and (6) investigate their stability. 

Ans. x = is a point of stable equilibrium; x = 4 is a point of unstable equilibrium 

2.78. Work Problem 2.77 if (a) V = a; 4 - 8a; 3 - 6a; 2 + 24a;, (6) V = x 4 . 

Ans. (a) x-\,2 are points of stable equilibrium; x — -1 is a point of unstable equilibrium. 
(6) x — is a point of stable equilibrium 

2.79. Work Problem 2.77 if V = sin 2^-a;. 

Ans. If n - 0, ±1, ±2, ±3, . . . then x = f + n are points of stable equilibrium, while x = J + n 
are points of unstable equilibrium. 

2.80. A particle moves in a force field with potential V = x 2 + y 2 + z 2 - Sx + 16y - 4z. Find the points 
of stable equilibrium. Ans. (4, —8, 2) 



MISCELLANEOUS PROBLEMS 

2.81. (a) Prove that F = {y 2 cos x + z s )i + (2y sin x - 4)j + (3a;z 2 + 2)k 
is a conservative force field. (6) Find the potential corresponding 
to F. (c) Find the work done in moving a particle in this field from 
(0,1,-1) to (tr/2, -1,2). 

Ans. (a) V = y 2 sin x + xz 3 - 4y + 2z + c, (6) 15 + 4n- 

2.82. A particle P is acted upon by 3 coplanar forces as indicated in 
Fig. 2-14. Find the force needed to prevent P from moving. 

Ans. 323 lb in a direction opposite to 150 lb force 

2.83. (a) Prove that F = r^x is conservative and (6) find the correspond- 
ing potential. Ans. (6) V = -^r 4 + c 



100 lb 




Fig. 2-14 



CHAP. 2] NEWTON'S LAWS OF MOTION. WORK, ENERGY AND MOMENTUM 59 

2.84. Explain the following paradox: According to Newton's third law a trailer pulls back on an auto- 
mobile to which it is attached with as much force as the auto pulls forward on the trailer. Therefore 
the auto cannot move. 

2.85. Find the potential of a particle placed in a force field given by F = — Kr~ n r where k and n are 
constants. Treat all cases. 

2.86. A waterfall 500 ft high has 440,000 ft 3 of water flowing over it per second. Assuming that the 
density of water is 62.5 lb/ft 3 and that 1 horsepower is 550 ft lb/sec, find the horsepower of the 
waterfall. Ans. 25 X 10 6 hp 

2.87. The power applied to a particle by a force field is given as a function of time t by <P(t) = 3* 2 — At + 2. 
Find the work done in moving the particle from the point where t = 2 to the point where t = 4. 
Ans. 36 

2.88. Can the torque on a particle be zero without the force being zero? Explain. 

2.89. Can the force on a particle be zero without the angular momentum being zero? Explain. 

2.90. Under the influence of a force field F a particle of mass 2 moves along the space curve 
r = 6t 4 i — 3* 2 j + (4t 3 — 5)k. Find (a) the work done in moving the particle from the point where 
t — to the point where t = l, (6) the power applied to the particle at any time. 

Ans. (a) 756 (6) 72*(48i 4 + St + 1) 

2.91. A force field moves a particle of mass m along the space curve r = acos«£i+ ftsinwi j. (a) What 
power is required? (6) Discuss physically the case a — b. Ans. (a) m(a 2 — 6 2 )« 3 sinwt coswi 

2.92. The angular momentum of a particle is given as a function of time t by 

12 = 6tH - (2t + l)j + (12i 3 - 8* 2 )k 
Find the torque at the time t = l. Ans. 12i — 2j + 20k 

2.93. Find the constant force needed to give an object of mass 36,000 lb a speed of 10 mi/hr in 5 minutes 
starting from rest. Ans. 1760 poundals 

2.94. A constant force of 100 newtons is applied for 2 minutes to a 20 kg mass which is initially at rest, 
(a) What is the speed achieved? (6) What is the distance traveled? 

Ans. (a) 600 m/sec, (6) 36,000 m 

2.95. A particle of mass m moves on the x axis under the influence of a force of attraction toward origin O 
given by F = — (k/x 2 )L If the particle starts from rest at x = a, prove that it will arrive at O in a 
time given by ^iray 'ma/2*. 

2.96. Work Problem 2.95 if F = -( K /x*)i. 

2.97. A particle of mass 2 units moves in the force field F = t 2 i — Btj + (t + 2)k where t is the time. 
(a) How far does the particle move from * = to t = 3 if it is initially at rest at the origin? 
(6) Find the kinetic energy at times t = 1 and t = 3. (c) What is the work done on the particle by 
the field from t = 1 to t = 3? (d) What is the power applied to the particle at * = 1? (e) What is the 
impulse supplied to the particle at t = 1 ? 

2.98. At f = 0a particle of unit mass is at rest at the origin. If it is acted upon by a force F = 100te _2t i, 
find (a) the change in momentum of the particle in going from time t = 1 to t = 2, (6) the velocity 
after a long time has elapsed. Ans. (a) 25e _2 (3 — 5e -2 )i, (6) 25 



60 NEWTON'S LAWS OF MOTION. WORK, ENERGY AND MOMENTUM [CHAP. 2 

2.99. A particle of mass 3 units moves in the xy plane under the influence of a force field having potential 

V = 6» 3 + 12y 3 + S6xy — 48s 2 . Investigate the motion of the particle if it is displaced slightly 
from its equilibrium position. 

[Hint. Near x = 0, y = the potential is very nearly 36xy - 48x 2 since 6a; 3 and 12y s are negligible.] 

2.100. A particle of unit mass moves on the x axis under the influence of a force field having potential 

V = 6x(x — 2). (a) Show that x = 1 is a position of stable equilibrium. (6) Prove that if the mass 
is displaced slightly from its position of equilibrium it will oscillate about it with period equal to 

4rv / 3. 

[Hint. Let x = 1 + u and neglect terms in u of degree higher than one.] 

2.101. A particle of mass m moves in a force field F = — kx\. (a) How much work is done in moving the 
particle from x = x x to x = x 2 ' (*») If a unit particle starts at x = x lt wit h speed v lt what is its speed 
on reaching x = x 2 1 Ans. (a) %k{x\ - x\), (6) V v 2 + (ic/m)(cc 2 - a; 2 ) 

2.102. A particle, of mass 2 moves in the xy plane under the influence of a force field having potential 
y = z 2 + y 2. The particle starts at time t = from rest at the point (2, 1). (a) Set up the differential 
equations and conditions describing the motion. (6) Find the position at any time t. (c) Find the 
velocity at any time t. 

2.103. Work Problem 2.102 if V - Bxy. 

2.104. Does Theorem 2.7, page 36, hold relative to a non-inertial frame of reference or coordinate system? 
Prove your answer. 

2.105. (a) Prove that if a particle moves in the xy plane under the influence of a force field having potential 

V = 12x(Sy - 4x), then x = 0, y = is a point of stable equilibrium. (6) Discuss the relationship 
of the result in (a) to Problem 2.37, page 53. 

2.106. (a) Prove that a sufficient condition for the point (a,b) to be a minimum point of the function 

V(x, y) is that at (a, b) 



(i) 



° V = d Z=0, (ii) . ^ (SrYSUS^ > and^>0 



to = W = "' w " VteVV* 8 / \ s »* 



(6) Use (a) to investigate the points of stability of a particle moving in a force field having potential 
y = jj.3 + y z - 33. _ \2y. Ans. (b) The point (1, 2) is a point of stability 

2.107. Suppose that a particle of unit mass moves in the force field of Problem 2.106. Find its speed 
at any time. 

2.108. A particle moves once around the circle r = a(cos B i + sin e j) in a force field 

F = (xi-yj)/(x 2 + y 2 ) 

(a) Find the work done. (6) Is the force field conservative? (c) Do your answers to (a) and (6) 
contradict Theorem 2.4, page 35? Explain. 

2.109. It is sometimes stated that classical or Newtonian mechanics makes the assumption that space and 
time are both absolute. Discuss what is meant by this statement. 

ft 2 

J Fdt 

2.110. The quantity F av = I 1 _ . is called the average force acting on a particle from time t x to t 2 . 

2 1 

Does the result (3) of Problem 2.5, page 40, hold if F is replaced by F av ? Explain. 



CHAP. 2] NEWTON'S LAWS OF MOTION. WORK, ENERGY AND MOMENTUM 61 

2.111. A particle of mass 2 gm moves in the force field F = Sxyi + (4x 2 — 8z)j — Syk. dynes. If it has 
a speed of 4 cm/sec at the point (—1, 2, —1), what is its speed at (1, —1, 1)? Ans. 6 cm/sec 

2.112. (a) Find positions of stable equilibrium of a particle moving in a force field of potential 

V = 18r 2 e-2r. 

(o) If the particle is released at r = £, find the speed when it reaches the equilibrium position, 
(c) Find the period for small oscillations about the equilibrium position. 

2.113. According to Einstein's special theory of relativity the mass m of a particle moving with speed 
v relative to an observer is given by m = m /Vl — v 2 /c 2 where c is the speed of light [186,000 
mi/sec] and m is the rest mass. What is the percent increase in rest mass of (a) an airplane moving 
at 700 mi/hr, (6) a planet moving at 25,000 mi/hr, (c) an electron moving at half the speed of light? 
What conclusions do you draw from these results? 



2.114. Prove that in cylindrical coordinates, 

dV ldV , dV 

dp e " + P 30 ** dz 
where e p , e^, e z are unit vectors in the direction of increasing p, <f> and z respectively. 



VF = ^-e p + i ^-e* + ^-e, 



2.115. Prove that in spherical coordinates, 

T7T/ dV . ! dV _l 1 3V 

Vy = ^ + r Jt e ° + ^In7"s7 e * 

where e r , e e , e^ are unit vectors in the direction of increasing r, e, <f> respectively. 






UNIFORM FORCE FIELDS 

A force field which has constant magni- 
tude and direction is called a uniform or con- 
stant force field. If the direction of this field 
is taken as the negative z direction as indi- 
cated in Fig. 3-1 and the magnitude is the 
constant F > 0, then the force field is given by 

F = -F k {1) 




urn 



-F«k 



Fig. 3-1 



UNIFORMLY ACCELERATED MOTION 

If a particle of constant mass m moves in a uniform force field, then its acceleration 

is uniform or constant. The motion is then described as uniformly accelerated motion. 

Using F = ma in (1), the acceleration of a particle of mass m moving in the uniform force 

field (1) is given by „ 

a = -^k (*) 

m v 



WEIGHT AND ACCELERATION DUE TO GRAVITY 

It is found experimentally that near the earth's 
surface objects fall with a vertical acceleration 
which is constant provided that air resistance is 
negligible. This acceleration is denoted by g and 
is called the acceleration due to gravity or the 
gravitational acceleration. The approximate mag- 
nitude of g is 980 cm/sec 2 , 9.80 m/sec 2 or 32 ft/sec 2 
according as the cgs, mks or fps system of units 
is used. This value varies at different parts of the 
earth's surface, increasing slightly as one goes 
from the equator to the poles. 

Assuming the surface of the earth is repre- 
sented by the xy plane of Fig. 3-2, the force acting 
on a particle of mass m is given by 



W = — mgk 



(3) 



This force, which is called the weight of the par- 
ticle, has magnitude W — mg. 




Fig. 3-2 



62 



CHAP. 3] MOTION IN A UNIFORM FIELD. FALLING BODIES AND PROJECTILES 63 

GRAVITATIONAL SYSTEM OF UNITS 

Because W = mg, it follows that m = Wig. This fact has led many scientists and engi- 
neers, who deal to a large extent with mechanics on the earth's surface, to rewrite the 
equations of motion with the fundamental mass quantity m replaced by the weight quantity 
W. Thus, for example, Newton's second law is rewritten as 

W 
F = fa W 

In this equation W and g can both vary while m = Wig is constant. One system of 
units used in (4) is the gravitational or English engineering system where the unit of F or 
W is the pound weight (lb wt) while length is in feet and time is in seconds. In this case 
the unit of m is the slug and the system is often called the foot-slug -second (fss) system. 
Other systems are also possible. For example, we can take F or W in kilograms weight 
(kg wt) with length in meters and time in seconds. 



ASSUMPTION OF A FLAT EARTH 

Equation (3) indicates that the force acting on mass m has constant magnitude mg and 
is at each point directed perpendicular to the earth's surface represented by the xy plane. 
In reality this assumption, called the assumption of the flat earth, is not correct first because 
the earth is not flat and second because the force acting on mass m actually varies with the 
distance from the center of the earth, as shown in Chapter 5. 

In practice the assumption of a flat earth is quite accurate for describing motions of 
objects at or near the earth's surface and will be used throughout this chapter. However, 
for describing the motion of objects far from the earth's surface the methods of Chapter 5 
must be employed. 



FREELY FALLING BODIES 

If an object moves so that the only force acting upon it is its weight, or force due to 
gravity, then the object is often called a freely falling body. If r is the position vector and 
m is the mass of the body, then using Newton's second law the differential equation of 
motion is seen from equation (3) to be 

d 2 r , d*r 

m dT> = ~ mgk or d¥ = ~ 9k (*) 

Since this equation does not involve the mass m, the motion of a freely falling body is 
independent of its mass. 



PROJECTILES 

An object fired from a gun or dropped from a moving airplane is often called a projectile. 
If air resistance is negligible, a projectile can be considered as a freely falling body so that 
its motion can be found from equation (5) together with appropriate initial conditions. If air 
resistance is negligible the path of a projectile is an arc of a parabola (or a straight line 
which can be considered a degenerate parabola). See Problem 3.6. 



64 MOTION IN A UNIFORM FIELD. FALLING BODIES AND PROJECTILES [CHAP. 3 

POTENTIAL AND POTENTIAL ENERGY 
IN A UNIFORM FORCE FIELD 

The potential of the uniform force field, or potential energy of a particle in this force 

field, is given by , v 

V = Fo(z-zo) (6) 

where z is an arbitrary constant such that when z = z , V = 0. We call z = z the reference 
level. 

In particular for a constant gravitational field, F = mg and the potential energy of the 

particle is 

V = mg(z-z ) v) 

This leads to 

Theorem 3.1. The potential energy of a particle in a constant gravitational field is 
found by multiplying the magnitude of its weight by the height above some prescribed 
reference level. Note that the potential energy is the work done by the weight in moving 
through the distance z — zo. 

MOTION IN A RESISTING MEDIUM 

In practice an object is acted upon not only by its weight but by other forces as well. 
An important class of forces are those which tend to oppose the motion of an object. Such 
forces, which generally arise because of motion in some medium such as air or water, are 
often called resisting, damping or dissipative forces and the corresponding medium is said 
to be a resisting, damping or dissipative medium. 

It is found experimentally that for low speeds the resisting force is in magnitude propor- 
tional to the speed. In other cases it may be proportional to the square [or some other power] 
of the speed. If the resisting force is R, then the motion of a particle of mass m in an 
otherwise uniform (gravitational) force field is given by 

dt 2 
If R = this reduces to (5). 



m-^ = mgk — R (#) 



ISOLATING THE SYSTEM 

In dealing with the dynamics or statics of a particle [or a system of particles, as we shall 
see later] it is extremely important to take into account all those forces which act on the 
particle [or on the system of particles]. This process is often called isolating the system. 

CONSTRAINED MOTION 

In some cases a particle P must move along some specified curve or surface as, for 
example, the inclined plane of Fig. 3-3 or the inner surface of a hemispherical bowl of 
Fig. 3-4 below. Such a curve or surface on which the particle must move is called a 
constraint and the resulting motion is called constrained motion. 

Just as the particle exerts a force on the constraint, there will by Newton's third law 
be a reaction force of the constraint on the particle. This reaction force is often described 
by giving its components N and f, normal to and parallel to the direction of motion 
respectively. In most cases which arise in practice, f is the force due to friction and is 
taken in a direction opposing the motion. 



CHAP. 3] MOTION IN A UNIFORM FIELD. FALLING BODIES AND PROJECTILES 



65 





n\ 


>&^ 




A 






s^ W 


s^\ a 






Fig. 3-3 



Fig. 3-4 



Problems involving constrained motion can be solved by using Newton's second law 
to arrive at differential equations for the motion and then solving these equations subject to 
initial conditions. 



FRICTION 

In the constrained motion of particles, one of the 
most important forces resisting motion is that due to 
friction. Referring to Fig. 3-5, let N be the magnitude 
of the normal component of the reaction of the con- 
straint on the particle m. Then it is found experi- 
mentally that the magnitude of the force f due to 
friction is given by 

/ = /JV (P) 




Fig. 3-5 



where /* is called the coefficient of friction. The direction of f is always opposite to the 
direction of motion. The coefficient of friction, which depends on the material of both 
the particle and constraint, is taken as a constant in practice. 



STATICS IN A UNIFORM GRAVITATIONAL FIELD 

As indicated in Chapter 2, a particle is in equilibrium under the influence of a system of 
forces if and only if the net force acting on it is F = 0. 



Solved Problems 



Fi 



UNIFORM FORCE FIELDS AND UNIFORMLY ACCELERATED MOTION 

3.1. A particle of mass m moves along a 
straight line under the influence of a con- 
stant force of magnitude F. If its initial 
speed is v , find (a) the speed, (b) the 
velocity and (c) the distance traveled 
after time t. Fig. 3-6 







66 MOTION IN A UNIFORM FIELD. FALLING BODIES AND PROJECTILES [CHAP. 3 

(a) Assume that the straight line along which the particle P moves is the x axis, as shown in 
Fig. 3-6 above. Suppose that at time t the particle is at a distance x from origin 0. If i is a 
unit vector in the direction OP and v is the speed at time t, then the velocity is vi. By Newton's 
second law we have 

4r(mvi) = Fi or m^j = F (1) 

dt dt 

Thus dv = —dt or J dv = J — dt 

i.e. v = —t+C! {2) 

m 

where c-^ is a constant of integration. To find c x we note the initial condition that v = v at 
t = so that from (2), c x = v and 

v = — t + v or v = v + — t (S) 

(6) From (5) the velocity at time t is 

If F 

vi = v i H ti or v = v H 1 

where v = vi, v = v i and F = Fi. 

(c) Since v = dse/d* we have from (8), 

* = * + £« » r *=(•, + £♦)* 
Then on integrating, assuming c 2 to be the constant of integration, we have 

* = * + (k)." + * 

Since a; = at t = 0, we find c 2 = 0. Thus 

• = * + (£)* w 

3.2. Referring to Pr oblem 3.1, sho w that the speed of the particle at any position x is 
given by v = V v l + (2F/m)x. 

Method 1. 

From (S) of Problem 3.1, we have t = m(v - v )/F. Substituting into (4) and simplifying, 
we find x = (m/2F)(v 2 - v„). Solving for v we obtain the required result. 

Method 2. 

From {1) of Problem 3.1, we have 

dv _ F . dv dx _ F 

dt ~ m' ' dx dt m 

or since v = dx/dt, 

„*1 = £, i. e . vdv = ?-dx 

dx m' m 

v 2 _ F , 
Integrating, y - J£* + c 3 



Since v = v when g = 0, we find c 3 = «g/2 and hence v = V«{j + (2F/m)x. 

Method 3. 

Change in kinetic energy from t = to any time t 

= Work done in moving particle from * = to any position x 



or $mv* - %mv% = F(x - 0). Then v = V^ + (2F/m)x. 



CHAP. 3] MOTION IN A UNIFORM FIELD. FALLING BODIES AND PROJECTILES 



67 



LINEAR MOTION OF FREELY 
FALLING BODIES 

3.3. An object of mass m is thrown vertically up- 
ward from the earth's surface with speed vo. 
Find (a) the position at any time, (b) the time 
taken to reach the highest point and (c) the 
maximum height reached. 

(a) Let the position vector of m at any time t be 
r = xi + yj + zk. Assume that the object starts 
at r = when t = 0. Since the force acting on 
the object is — mgk, we have by Newton's law, 



F = 




Fig. 3-7 



d 2 r 
dt 2 



dv . 

m-r- = — mgk 



dv 
dt 



= -fl* 



(J) 



where v is the velocity at time t. Integrating (1) once yields 

v = -gtk + c t (2) 

Since the velocity at t = [i.e. the initial velocity] is v<jk> we have from (2), e t = Vok so that 

v = — gtk + V(jk = (v — gt)k 



(S) 
(*) 

(5) 

(6) 
(7) 



or -^ = (v -gt)k 

Integrating (4) yields r = (v t — %gt 2 )k + c 2 

Then since r = when t = 0, c 2 = 0. Thus the position vector is 

r = (v Q t-$gt2)k 
or, equivalently, x = 0, y = 0, z = v t — ^gt 2 

(6) The highest point is reached when v = (v — gt)k = 0, i.e. at time t = vjg. 
(c) At time t = VfJg the maximum height reached is, from (7), z = i%/2g. 

Another method. 

If we assume, as is physically evident, that the object must always be on the z axis, we may 
avoid vectors by writing Newton's law equivalently as [see equation (1) above and place r = zk] 

cPz/dt 2 = -g 

from which, using z = 0, dz/dt = v at t = 0, we find 

z = v t - %gt 2 

as above. The answers to (6) and (c) are then obtained as before. 



3.4. Find the speed of the particle of Problem 3.3 in terms of its distance from origin O. 
Method 1. From Problem 3.3, equations (3) and (7), we have 

v = v - gt, z = v t — \gt 2 
Solving for t in the first equation and substituting into the second equation, we find 



2 _ «2 



/v -v\ /v -v\ 2 v^-v 

= Vo \-T-)-t g (-T-) = -*- 



„2 — „2 _ 



2gz 



Method 2. From equation (1) of Problem 3.3 we have, since v = vk and v = dz/dt, 
dv dv dz __. .. dv 

dt 



dv dz _ 

dz dt ~ 9 



V Tz = ~ g 



= — g, i.e. — — = — g or 

Then on integrating, vV2 = — gz + c 3 . Since v = v at z = 0, c 3 = v^/2 and thus v 2 = v 2 — 2gz. 
Method 3. See Problem 3.9 for a method using the principle of conservation of energy. 



68 



MOTION IN A UNIFORM FIELD. FALLING BODIES AND PROJECTILES [CHAP. 3 



MOTION OF PROJECTILES 

3.5. A projectile is launched with initial speed 
v at an angle a with the horizontal. Find 
(a) the position vector at any time, (b) the 
time to reach the highest point, (c) the 
maximum height reached, (d) the time of 
flight back to earth and (e) the range. 

(a) Let r be the position vector of the projec- 
tile and v the velocity at any time t. Then 
by Newton's law 




d 2 r , 



i.e., 



dt 2 



(1) 
= -flrk 



Fig. 3-8 



dv , 



(2) 



Integration yields 



v = -gtk + c x W 
Assume the initial velocity of the projectile is in the yz plane so that the initial velocity is 

v = v cos a j + v sin a k (4) 
Since v = v at t = 0, we find from (S), 

v = v cos a j + (v sin a — gt)k (5) 
Replacing v by dr/dt in (5) and integrating, we obtain 

r = (v cos a)tj + {(v Q sin a)t- %gt 2 }k (6) 

or, equivalents, x = 0, # = (v cos a)*, 2 = (v sin a)* - \g& (7) 

It follows that the projectile remains in the yz plane. 

(b) At the highest point of the path the component of velocity v in the k direction is zero. Thus 

Vq sin a 



is the required time. 



v Q sin a 



gt = and t = 



g 



(c) Using the value of t obtained in (&), we find from (7) that 

'v sin a 



Maximum height reached = (v sin a) 



9 



-y 



t> sina\ 2 aj^sin 2 a 



2g 



(8) 



(9) 



(d) The time of flight back to earth is the time when z = 0, i.e. when 

{v sin a)t - y& = t[(v Q sin a) - ±gt] = 
or since t ¥* 0, 



t = 



2v sin a 



<i0) 



Note that this is twice the time in (6). 
(e) The range is the value of y at the time given by (10), i.e., 



/ 2v sin a \ _ 
Range = (v cos a) ( ) - 



2v sin a cos a 



vl sin 2a 



3.6. Show that the path of the projectile in Problem 3.5 is a parabola. 

From the second equation of (7) in Problem 3.5, we have t = y/(v cosa). Substituting this 
into the third equation of (7) in Problem 3.5, we find 

z = {v sin a)(y/v cos a) - y{yh cos a) 2 or z = y tan « - (f//2 V 2 )2/ 2 sec 2 a 

which is a parabola in the yz plane. 



CHAP. 3] MOTION IN A UNIFORM FIELD. FALLING BODIES AND PROJECTILES 



69 



3.7. Prove that the range of the projectile of Problem 3.5 is a maximum when the launch- 
ing angle a = 45°. 

By Problem 3.5(e) the range is (v§ sin 2a)/ g. This is a maximum when sin 2a = 1, i.e. 2a = 90° 
or a = 45°. 



POTENTIAL AND POTENTIAL ENERGY 
IN A UNIFORM FORCE FIELD 

3.8. (a) Prove that a uniform force field is conservative, (b) find the potential correspond- 
ing to this field and (c) deduce the potential energy of a particle of mass m in a 
uniform gravitational force field. 

(a) If the force field is as indicated in Fig. 3-1, then F = — F k. We have 



V XF = 



Thus the force field is conservative. 



i J k 

d/dx d/dy d/dz 
-F 



= 



dV. dV . dV. „, dV . dV dV „ . . . , 

— — i — — j - — k. Then — =0, — = 0, — = F from which 
dx dy dz dx dy dz 



(b) F = -F k = -VV 

V = F Q z + c. If V = at z = z , then c = —F z and so V = F (z — z ) 



(c) For a uniform gravitational force field, F = -mgls. [see Fig. 3-2, page 62] and corresponds 
to F = mg. Then by part (b) the potential or potential energy is V = mg(z — z Q ). 



3.9. Work Problem 3.4 using the principle of conservation of energy. 
According to the principle of conservation of energy, we have 

P.E. at z = + K.E. at z = = P.E. at z + K.E. at z 
^mvl = mgz + |mi) 2 



+ 



Then v 2 = v* - 2gz. 



MOTION IN A RESISTING MEDIUM 

3.10. At time t = a. parachutist [Fig. 3-9] having 
weight of magnitude mg is located at z = and 
is traveling vertically downward with speed vo. 
If the force or air resistance acting on the 
parachute is proportional to the instantaneous 
speed, find the (a) speed, (b) distance traveled 
and (c) acceleration at any time t > 0. 

(a) Assume the parachutist (considered as a particle 
of mass m) is located at distance z from origin O. 
If k is a unit vector in the vertically downward 
direction, then the weight is mgls. while the force 
of air resistance is — (3vk so that the net force 
is (mg — fiv)k. 



Thus by Newton's law, 
(mg 



ra-r-k = 
at 




/3v)k 



(1) 



Fig. 3-9 



70 MOTION IN A UNIFORM FIELD. FALLING BODIES AND PROJECTILES [CHAP. 3 

dv „ mdv 

i.e. wi-jt = wgr — Bv or r- = dt 

dt mg — Bv 

Integrating, - 7f ln ( mg ~ Bv ^ ~ l + c i (*) 

AM 

Since v = v at t = 0, c x = — — In (ra^r — Bv ). Then from (2), 

t = jln(mg-pv ) -jln (mg-/3v) = jm ( ^_^ j 

Thus X-7* = e " t/m or * = T + ( v °~ 2 f) e ~ &t/m {s) 

(b) From (5), dz/dt = mg/B + (v — mg/B)e-W m . Then by integration, 

mpt m/ mg\ _ Mlm . 
z = -f- -jf^v --^je 3t/m + C2 

Since z = at £ = 0, c 2 = (m/B)(v — mg/B) and thus 

« = ^+7 (•*-?) a -•"•"-) 

(c) From (5), the acceleration is given by 



04) 



• = % = -£(»»-f>- e " m = ('-S)'-"- <*> 

3.11. Show that the parachutist of Problem 3.10 approaches a limiting speed given by mglp. 

Method 1. 

From equation (5) of Problem 3.10, v = mg/B + (v — mg/B)e-W m . Then as t increases, 
v approaches mg/B so that after a short time the parachutist is traveling with speed which is 
practically constant. 

Method 2. 

If the parachutist is to approach a limiting speed, the limiting acceleration must be zero. 
Thus from equation (1) of Problem 3.10 we have mg — Bv hm = or v lim = mg//3. 

3.12. A particle of mass m is traveling along the x axis such that at t — it is located at x = 
and has speed vo. The particle is acted upon by a force which opposes the motion 
and has magnitude proportional to the square of the instantaneous speed. Find the 
(a) speed, (b) position and (c) acceleration of the particle at any time t > 0. 

(a) Suppose particle P is at a distance x from O at 

* = and has speed v [see Fig. 3-10]. Then the F = -Bv 2 i 

force F = — Bv 2 i where B > is a constant of 
proportionality. By Newton's law, 

m-rri = —Bv 2 i or -5- = — —dt (J) 

dt ^ v 2 m 



Integrating, — 1/v = —Bt/m + c v Since v = v 

when t = 0, we have c x = -1/iv Thus Fig. 3-10 

1 /3t 1 wv 

= — or v = — — t— r — 

v m v Bv t + m 

which is the speed. 
(6) From W, f = .r? . Then f i. = f „-£*-* = =» f ■ 



dt 



Bv t + m ' J J Bv t + m Bv J t + m/Bv 

m ^ ( * , »\ , 



— or 



CHAP. 3] MOTION IN A UNIFORM FIELD. FALLING BODIES AND PROJECTILES 71 

Since x = at t = 0, c 2 = -^ln(-^V Thus 

• = M (+ ^)-f ln te) = f'»( i+ ^) 

(c) From (a), 

_ £k _ d ( mv ° \ = PmVQ u\ 

a ~ dt dt\pv t + mj (f3v t + m)* w 

Note that although the speed of the particle continually decreases, it never conies to rest. 



3.13. Determine the (a) speed and (b) acceleration of the particle of Problem 3.12 as a 
function of the distance x from O. 

Method 1. From parts (a) and (6) of Problem 3.12, 

m /fiv t + m\ mv /3v t + m v 

In , and " — ~ — 



P \ m ) ' fiv Q t + m m v 

Then s=jln(^J or v = i> «-** /m 

and the acceleration is given in magnitude by 

dv Pn _ Br/m dx P v l _ 2Bx/m 

dt m dt m 

which can also be obtained from equation (4.) of Problem 3.12. 

Method 2. From equation (1) of Problem 3.12 we have 

dv dv dx dv n „ 

m-77 = m-j- -77 = mv-j- = — Bv 2 
dt dx dt dx 

or since v ¥= 0, m -=— = — /8v and — = — —x. Integrating, lnv = —fix/m + c 3 . Since v = v 
ax v m 

when x = 0, c 3 = In v . Thus In (v/v ) = —Bx/m or v = VQe - ^/™. 



3.14. Suppose that in Problem 3.5 we assume that the projectile has acting upon it a force 
due to air resistance equal to — pv where /? is a positive constant and v is the instan- 
taneous velocity. Find (a) the velocity and (b) the position vector at any time. 

(a) The equation of motion in this case is 

cPt dv 

m-j-g = — mgk — /?v or m ~Jt^~ P y = — w #k C) 

( 0/m dt 

Dividing by m and multiplying by the integrating factor e J = eP t/m , the equation can 

be written as 

±{ e fit/m y } = -gefit/m^ 

Integration yields e Pt/m v = — ■£- e& t/m k + c x (2) 

The initial velocity or velocity at t = is 

Vo = v Q cos a j + v Q sin a k (8) 

Using this in (2) we find 

c i = v o c °s o] + v<> sin a k -I — ^-k 



72 



MOTION IN A UNIFORM FIELD. FALLING BODIES AND PROJECTILES [CHAP. 3 



Thus (2) becomes on dividing by e^ t/m , 



v = (v cos a j + v sin a k)e~^ m — ^(1 — e~W™)k 



(b) Replacing v by dr/dt in (4) and integrating, we find 



m 9 <* jl m — < 



Since r = at * = 0, 



Using (6) in (5), we find 
mv Q 



r = ~"g-( v 0COSa j + v sinak)e-^"» f-(t + ^ e -flt/m)k + Ca 



fi v /? 






J8 2 ' 



r = -^(cosa j + sin a k)(l - e-"t/m) - 2M / t + ™ -/3t/m _ 2?M k 



(4) 

(5) 
(*) 
(7) 



3.15. Prove that the projectile of Problem 3.14 attains a limiting velocity and find its value. 

Method 1. 

Refer to equation (4) of Problem 3.14. As t increases, e~» t/m approaches zero. Thus the 
velocity approaches a limiting value equal to v llm = —(mg/(3)k. 

Method 2. 

If the projectile is to approach a limiting velocity its limiting acceleration must be zero. Thus 
from equation (1) of Problem 3.14, -mgk - 0v llm = or v lim = -(mg/p)k. 



CONSTRAINED MOTION 

3.16. A particle P of mass m slides without rolling 
down a f rictionless inclined plane AB of angle a 
[Fig. 3-11]. If it starts from rest at the top A 
of the incline, find (a) the acceleration, (b) the 
velocity and (c) the distance traveled after 
time t. 

(a) Since there is no friction the only forces acting 
on P are the weight W = — mgk and the re- 
action force of the incline which is given by the 
normal force N. 

Let e x and e 2 be unit vectors parallel and 
perpendicular to the incline respectively. If we 
denote by s the magnitude of the displacement 
from the top A of the inclined plane, we have 
by Newton's second law 



— * 


N / 


N. pjf /■ inff si" a «i 




<^. 


W = -mgk 
— mg cos a e 2 





Fig. 3-11 



d 2 
m-j-zisei) = W + N = raflrsinae! 



<J> 



since the resultant equal to W + N is mg sin a e^ as indicated in Fig. 3-11. From (I) we have 

d?s/d& - g sin a (2) 

Thus the acceleration down the incline at any time t is a constant equal to g sin a. 

(6) Since v = ds/dt is the speed, (2) can be written 

dv/dt = flrsina or v — (g sin a)t + c t 

on integrating. Using the initial condition v = at t = 0, we have c x = so that the 

speed at any time t is 

v = (g sin a)t (3) 

The velocity is ve t = (g sin ajte^ which has magnitude (g sin a)t in the direction e t down 
the incline. 



CHAP. 3] MOTION IN A UNIFORM FIELD. FALLING BODIES AND PROJECTILES 



73 



(c) Since v = ds/dt, (5) can be written 

ds/dt = (g sin a)t or s = %(9 sin a)t 2 + c 2 

on integrating. Using the initial condition s = at t = 0, we find c 2 = so that the 

required distance traveled is ,, . w , /»\ 

8 = -|(p sin a)t 2 W 

3.17. If the length AB of the incline in Problem 3.16 is I, find (a) the time r taken for the 
particle to reach the bottom B of the incline and (&) the speed at B. 

(a) Since 8 = 1 at B, the time t to reach the bottom is from equation (4) of Problem 3.16 given by 
I = ^(g sin a )r 2 or r - }/2l/(g sin a). 

(6) The speed at £ is given from (5) of Problem 3.16 by v = (g sin «)r = V2tf* sin a. 



MOTION INVOLVING FRICTION 

3.18. Work Problem 3.16 if the inclined plane has 
a constant coefficient of friction ju.. 

(a) In this case there is, in addition to the forces W 
and N acting on P, a frictional force f [see Fig. 
3-12] directed up the incline [in a direction oppo- 
site to the motion] and with magnitude 

liN = v.mg cos a CO 



i.e. 



pmg cos a 
i = —ping cos a ^ 



(2) 





N> 


f f s mg sin a ti 




vjW 


W = -mgk 


' s \ / e 2 


*) ^>V 


— mg cos a e 2 ' a /\^ 



Fig. 3-12 



Then equation (1) of Problem 3.16 is replaced by 

cP(*e t ) 

= W + N + f = mflr sin a x e x — y.mg cos a ej 



m- 



dt 2 



(S) 



or <Ps/dt* = flr(sin a - /* cos a) W 

Thus the acceleration down the incline has the constant magnitude g(sin a - ? cos a) provided 
sin a > /icosa or tana > /* [otherwise the frictional force is so great that the particle will 
not move at all]. 

(6) Replacing d 2 s/dt 2 by dv/dt in U) and integrating as in part (6) of Problem 3.16, we find the 

speed at any time t to be (K x 

v = gr( S in a — /t COS a)t l°J 



(c) Replacing v by <*«/<** in (5) and integrating as in part (c) of Problem 3.16, we find 

8 = ^g(sina — (i cos a)t 2 



(6) 



3.19. An object slides on a surface of ice along the horizontal straight line OA [Fig. 3-13]. 
At a certain point in its path the speed is v and the object then comes to rest after 
traveling a distance xo. Prove that the coefficient of friction is v 2 /2gx . 



Let * be the instantaneous distance of the 
object of mass m from O and suppose that at 
time t = 0, x = and dx/dt = v . 

Three forces act on the object, namely (1) the 
weight W = mg, (2) the normal force N of the 
ice surface on the object, and (3) the frictional 
force f. 



O 






mg 



Fig. 3-13 



By Newton's second law we have, if v is the instantaneous speed, 



w ^i = W + N + f 
dt 



(1) 



74 



MOTION IN A UNIFORM FIELD. FALLING BODIES AND PROJECTILES [CHAP. 3 



But since N = -W and the magnitude of f is f = pN = ymg so that £ = —fimgi, (1) becomes 

dv 
dt 



dv . . dv 

m-^i = -fimgi or -^ = -fig 



Method 1. Write (2) as 



dv dx dv 

dx"di = -m or v Tx = -** 



Then v dv = —figdx 

Integrating, using the fact that v = v at x = 0, we find 

v*/2 = -mx + v\l1 
Then since v = when x = x , (4) becomes 

-figx Q + v*/2 = or n = v*/2gx Q 



(-4) 
(5) 



Method 2. From (2) we have, on integrating and using the fact that v = v at t = 0, 

v = v — figt or dx/dt = v — figt (6) 

Integrating again, using the fact that x = at t = 0, we find 

x = v t- frgfi (7) 

From (7) we see that the object comes to rest (i.e., v = 0) when 

v — ngt = or t — v //xg 
Substituting this into (7) and noting that x = x , we obtain the required result. 



STATICS IN A UNIFORM GRAVITATIONAL FIELD 

3.20. A particle of mass m is suspended in equilibrium by two inelastic strings of lengths 
a and b from pegs A and B which are distant c apart. Find the tension in each string. 




L 




— mgk 



Fig. 3-14 



Fig. 3-15 



Let W denote the weight of the particle and T t and T 2 the respective tensions in the strings 
of lengths a and b as indicated in Fig. 3-14. These forces are also indicated in Fig. 3-15 and are 
assumed to lie in the plane of unit vectors j and k. By resolving T 1 and T 2 into horizontal and 
vertical components it is clear that 



= T x sin a k — T x cos a j, 



T 2 = T 2 sin p k + T 2 cos ft j 



where T 1 and T 2 are the magnitudes of T x .and T 2 respectively and where a and /J are the respective 
angles at A and B. Also we have 

W = — mgk. 

Since the particle is in equilibrium if and only if the net force acting on it is zero, we have 

F = T x + T 2 + W 

= 2 1 ! sin a k — T x cos a j + T 2 sin /? k + T 2 cos /? j — mgk 

= (T 2 cos p — T t cos a)j + (2\ sin a + T 2 sin p - mg)k 

= 



CHAP. 3] MOTION IN A UNIFORM FIELD. FALLING BODIES AND PROJECTILES 



75 



From this we must have 

T 2 cos /3 - T t cos a - 0, T x sin a + T 2 sin ft - mg = 

Solving simultaneously, we find 

„ _ mg cos p j, _ my cos a 

1 ~ sin (<* + £)' 2 sin (<* + /?) 

The angles a and /? can be determined from the law of cosines as 



a = COS -1 



a 2 + c 2 - b 2 
2ac 



b 2 + c 2 - a 2 
26c 



From these the tensions can be expressed in terms of a, b, c. 



MISCELLANEOUS PROBLEMS 

3.21. An inclined plane [Fig. 3-16] makes an 
angle a with the horizontal. A projectile is 
launched from the bottom A of the incline 
with speed vo in a direction making an 
angle jB with the horizontal. 

(a) Prove that the range R up the incline is 
given by 

2v 2 sin (/? — a) cos /? 



R = 




g COS z a 



Fig. 3-16 



(b) Prove that the maximum range up the incline is given by 



■Umax — 



and is achieved when /? = tt/4 + a/2. 



g(l + sin a) 



(a) As in Problem 3.5, equation (6), the position vector of the projectile at any time t is 

r = (v cos/?)«j + {(v Q sin p)t - lgt 2 }k (1) 

or y = (v cos /?)*, z = (v sin /?)£ - \gt 2 (2) 

The equation of the incline [which is a line in the yz plane] is 

z = y tan a (3) 

we see that the projectile's path and the incline intersect for those 
(v sirt /3)t — \gt 2 — [(v cos (3)t] tana 

2v (sin /? cos a — cos /3 sin a) 2v sin (/J — a) 



Using equations (2) in 
values of t where 



t = 



and 



g cos a 



g cos a 



The value t = gives the intersection point A. The second value of t yields point B 
which is the required point. Using this second value of t in the first equation of (2), we find 
that the required range R up the incline is 



R = y sec a = {v cos /3) 



2v sin (/? — o] 

flf COS a 



2v sin (/3 — a) cos /? 
cos 2 a 



(b) Method 1. The range R can be written by using the trigonometric identity 

sin A cos B = £{sin (A + B) + sin (A - B)} 



R = 



vl 



g cos 2 a 



{sin (2/3 — a) — sin a} 



76 



MOTION IN A UNIFORM FIELD. FALLING BODIES AND PROJECTILES [CHAP. 3 



This is a maximum when sin (2/3 - o) = 1, i.e. 20 - a = T /2 or fi = a/2 + tt/4, and the 
value of this maximum is 



R 



v, 



g cos z a 



(1 — sin a) = 



2 



#(1 — sin 2 a) 



(1 — sin a) — 



g(l + sin a) 



Method 2. 

The required result can also be obtained by the methods of differential calculus for 
finding maxima and minima. 



3.22. Two particles of masses mi and m 2 respectively are 
connected by an inextensible string of negligible mass 
which passes over a fixed frictionless pulley of negli- 
gible mass as shown in Fig. 3-17. Describe the motion 
by finding (a) the acceleration of the particles and 
(b) the tension in the string. 

Let us first isolate mass m,. There are two forces acting 
on it: (1) its weight m y g = m x gk, and (2) the force due to 
the string which is the tension T = — Tk. If we call a = dk 
the acceleration, then by Newton's law 

m x ak = m t gk — Tk (1) 

Next we isolate mass m 2 . There are two forces acting 
on it: (1) its weight m 2 g = m^k, and (2) the tension 
T = — Tk [the tension is the same throughout the string 
since the mass of the string is assumed negligible and in- 
extensible]. Since the string is inextensible, the acceleration 
of ra 2 is —a = — ak. Then by Newton's law 

— m 2 ak = m 2 gk — Tk (2) 



'mWWXW 



Wig 



m 2 



m 2 g 



Fig. 3-17 



From (1) and (2) we have 

m^a 

Solving simultaneously, we find 



m x g — T, 



-m 2 a 



m 2 g 



m i ~ m 2 
mi + m 2 



g, 



T = 



2m 1 m 2 
m 1 + m 2 i 



Thus the particles move with constant acceleration, one particle rising and the other falling. 

In this pulley system, sometimes called Atwood's machine, the pulley can rotate. However, 
since it is frictionless and has no mass [or negligible mass] the effect is the same as if the string 
passed over a smooth or frictionless peg instead of a pulley. In case the mass of the pulley 
is not negligible, rotational effects must be taken into account and are considered in Chapter 9. 



3.23. A particle P of mass m rests at the top A of a 
frictionless fixed sphere of radius b. The par- 
ticle is displaced slightly so that it slides (with- 
out rolling) down the sphere, (a) At what posi- 
tion will it leave the sphere and (b) what will 
its speed be at this position? 

The particle will slide down a circle of radius a 
which we choose to be in the xy plane as indicated in 
Fig. 3-18. The forces acting on the particle are: 
(1) its weight W = —mgj, and (2) the reaction force 
N of the sphere on the particle normal to the sphere. 

Method 1. 

(a) Let the position of the particle on the circle be 
measured by angle and let v i and t be unit 
vectors. Resolving W into components in direc- 
tions r x and $ lt we have as in Problem 1.43, 
page 24, 




Fig. 3-18 



CHAP. 3] MOTION IN A UNIFORM FIELD. FALLING BODIES AND PROJECTILES 77 

W = (W'r,)r, + (W •#!)#, 

= (— wgrj • r^ + (— mg\ • f x)*! = -mg sin o r x — mg cos ^ 

Also, N = Nt x 

Using Newton's second law and the result of Problem 1.49, page 26, we have 
F = ma = m[(r — re 2 )r x + (r'o + 2re)$ x ] 

= W + N = (2V — mflr sin tf)^ — mg cose 9 X (1) 

Thus ra(r-r« 2 ) = N-mg sine, m(r'e + 2r'e) = -mg cos e (2) 

While the particle is on the circle (or sphere), we have r = b. Substituting this into (2), 

—mbe 2 = N — mg sin 0, be = —g cos e (3) 

Multiplying the second equation by e, we see that it can be written 

d /e 2 \ d 

b Tt\2j = -*S (Bin#) 

Integrating, be 2 /2 = — g sin + c 1# Now when * = tt/2, = so that c x = g and 

6ff2 = 2gr(l - sin a) (4) 

Substituting (4) into the first equation of (S), we find 

N = mg(Z sin e - 2) (5) 

Now as long as N > the particle stays on the sphere; but when N = the particle will 
be just about to leave the sphere. Thus the required angle is given by 3 sin e — 2 = 0, i.e., 

sin* = 2/3 or e = sin" 1 2/3 (6) 

(6) Putting sin 6 = £ into (4), we find 

e 2 = 2^/36 (7) 

Then if v is the speed, we have v = be so that (7) yields v 2 = f&p or v = yj\bg. 

Method 2. By the conservation of energy, using the x axis as reference level, we have 
P.E. at A + K.E. at A = P.E. at P + K.E. at P 
mgb +0 = mgb sin e + fymv 2 

or v 2 = 2gb(l - sin e) («) 

Using the result of Problem 1.35, page 20, together with Newton's second law, we have, since 
the radius of curvature is 6, 

» = «• = Cf.-I'-) = w+n 

= (N — mg sin e)r x — mg cos e 6 X 

Using only the r t component, we have 

v 2 /b = N — mg sine (5) 

From (8) and (9) we find N = mg(3 sin e - 2) which yields the required angle sin" 1 (§) as in 
Method 1. The speed is then found from (8). 



78 MOTION IN A UNIFORM FIELD. FALLING BODIES AND PROJECTILES [CHAP. 3 

Supplementary Problems 

UNIFORM FORCE FIELDS AND LINEAR MOTION OF 
FREELY FALLING BODIES 

3.24. An object of mass m is dropped from a height H above th e grou nd. Prove that if air resistance 
is negligible, then it will reach the ground (a) in a time y/2H/g and (b) with speed y/2gH. 

3.25. Work Problem 3.24 if the object is thrown vertically downward with an initial velocity of magni- 
tude v . Ans. (a) (y/v* + 2gH - v )/g, (b) y/v* + 2gH 

3.26. Prove that the object of Problem 3.3, page 67, returns to the earth's surface (a) with the same 
speed as the initial speed and (6) in a time which is twice that taken to reach the maximum height. 

3.27. A ball which is thrown upward reaches its maximum height of 100 ft and then returns to the 
starting point, (a) With what speed was it thrown? (6) How long does it take to return? 

Ans. (a) 80 ft/sec, (6) 5 sec 

3.28. A ball which is thrown vertically upward reaches a particular height H after a time t x on the 
way up and a time t 2 on the way down. Prove that (a) the initial velocity with which the ball was 
thrown has magnitude $g(r t + t 2 ) and (6) the height H = $griT 2 . 

3.29. In Problem 3.28, what is the maximum height reached? Ans. ^g(r t + t 2 ) 2 

3.30. Two objects are dropped from the top of a cliff of height H. The second is dropped when the first 
has traveled a distance D. Prove that at the instant w hen t he first object has reached the bottom, 
the second object is at a distance above it given by 2yDH — D. 

3.31. An elevator starts from rest and attains a speed of 16 ft/sec in 2 sec. Find the weight of a 
160 lb man in the elevator if the elevator is (a) moving up (6) moving down. 

Ans. (a) 200 lb, (6) 120 lb 

3.32. A particle of mass 3 kg moving in a straight line decelerates uniformly from a speed of 40 m/sec 
to 20 m/sec in a distance of 300 m. (a) Find the magnitude of the deceleration. (6) How much 
further does it travel before it comes to rest and how much longer will this take? 

Ans. (a) 2 m/sec 2 , (6) 100 m; 10 sec 

3.33. In Problem 3.32, what is the total work done in bringing the particle to rest from the speed 
of 40 m/sec? Ans. 2400 newton meters (or joules) 

MOTION OF PROJECTILES 

3.34. A projectile is launched with a muzzle velocity of 1800 mi/hr at an angle of 60° with a 
horizontal and lands on the same plane. Find (a) the maximum height reached, (6) the time 
to reach the maximum height, (c) the total time of flight, (d) the range, (e) the speed after 
1 minute of flight, (/) the speed at a height of 32,000 ft. 

Ans. (a) 15.5 mi, (6) 71.4 sec, (c) 142.8 sec, (d) 35.7 mi, (e) 934 mi/hr, (/) 1558 mi/hr 

3.35. (a) What is the maximum range possible for a projectile fired from a cannon having muzzle 
velocity 1 mi/sec and (6) what is the height reached in this case? 

Ans. (a) 165 mi, (6) 41.25 mi 

3.36 A cannon has its maximum range given by R max . Prove that (a) the height reached in such case 
is £# max and (&) the time of flight is ^R m& J2g. 

3.37. It is desired to launch a projectile from the ground so as to hit a given point on the ground 
which is at a distance less than the maximum range. Prove that there are two possible angles 
for the launching, one which is less than 45° by a certain amount and the other greater than 
45° by the same amount. 

3.38. A projectile having horizontal range R reaches a maximum height H . Prove that it must have 
been launched with (a) an initial speed equal to y/g(R 2 + 16H 2 )/8H and (6) at an angle with 
the horizontal given by sin -1 (4H/y/R 2 + 16Jf? 2 ). 




CHAP. 3] MOTION IN A UNIFORM FIELD. FALLING BODIES AND PROJECTILES 79 

3.39. A projectile is launched at an angle a from a 
cliff of height H above sea level. If it falls into 
the sea at a distance D from the base of the 
cliff, prove that its maximum height above sea 

level is „ _,_ Ctan'q 

H+ 4(H + Ptan«) Fl S- 3 " 19 

MOTION IN A RESISTING MEDIUM 

3.40. An object of weight W is thrown vertically upward with speed v . Assuming that air resistance 
is proportional to the instantaneous velocity and that the constant of proportionality is k, prove 
that (a) the object will reach a maximum height of 

Wkv W 2 ln ( 1 + KV o 



K 2 g K 2 g \ W 
and that (6) the time taken to reach this maximum height is 

Kg \ W ) 

3.41. A man on a parachute falls from rest and acquires a limiting speed of 15 mi/hr. Assuming 
that air resistance is proportional to the instantaneous speed, determine how long it takes to reach 
the speed of 14 mi/hr. Arts. 1.86 sec 

3.42. A mass m moves along a straight line under the influence of a constant force F. Assuming 
that there is a resisting force numerically equal to kv 2 where v is the instantaneous speed and k 

w . /F-kv\ 



is a constant, prove that the distance traveled in going from speed v x to v 2 is ^ In ( F _ ^ 

3.43. A particle of mass m moves in a straight line acted upon by a constant resisting force of magni- 

tude F. If it starts with a speed of v , (a) how long will it take before coming to rest and 
(6) what distance will it travel in this time? Ana. (a) mv /F, (b) mv 2 /2F 

3.44. Can Problem 3.43 be worked by energy considerations? Explain. 

3.45. A locomotive of mass m travels with constant speed v along a horizontal track, (a) How long 
will it take for the locomotive to come to rest after the ignition is turned off, if the resistance 
to the motion is given by a + /3v 2 where v is the in stant aneous speed and a and /3 are constants? 
(b) What is the distance traveled? Ana. (a) y/mlp tan" 1 (v Q VJ/Z), (6) (m/2/3) In (1 + prf/a) 

3.46. A particle moves along the x axis acted upon only by a resisting force which is proportional 
to the cube of the instantaneous speed. If the initial speed is v and after a time r the speed is 
£v , prove that the speed will be ±v in time 5t. 

3.47. Find the total distance traveled by the particle of Problem 3.46 in reaching the speeds (a) |v > 
(6) £v . Ana. (a) §v r, (6) iv 

3.48. Prove that for the projectile of Problem 3.14, page 71, 

m f /?v sina\ 
(a) the time to reach the highest point is -j In ( 1 H — — ) , 

mv sina m 2 g , / J»oraa\ 
(6) the maximum height is tj p- In I 1 H — ) . 

CONSTRAINED MOTION AND FRICTION 

3.49. A weight of 100 lb slides from rest down a 60° incline of length 200 ft starting from the top. 
Neglecting friction, (a) how long will it take to reach the bottom of the incline and (6) what is 
the speed with which it reaches the bottom? Ana. (a) 3.80 sec, (&) 105.3 ft/sec 



80 



MOTION IN A UNIFORM FIELD. FALLING BODIES AND PROJECTILES [CHAP. 3 



3.50. Work Problem 3.49 if the coefficient of friction is 0.3. Ans. (a) 4.18 sec, (6) 95.7 ft/sec 

3.51. (a) With what speed should an object be thrown up a smooth incline of angle a and length I, 
starting from the bottom, so as to just reach the top and (6) what is the time taken? 

Ans. (a) y/2gl sin a, (b) s/2l/(g sin a) 

3.52. If it takes a time t for an object starting from speed v on an icy surface to come to rest, prove 
that the coefficient of friction is v Q /gr. 

3.53. What force is needed to move a 10 ton truck with uniform speed up an incline of 30° if the 
coefficient of friction is 0.1 1 Ans. 5.87 tons 

3.54. A mass m rests on a horizontal piece of wood. The wood is tilted upward until the mass m just 
begins to slide. If the angle which the wood makes with the horizontal at that instant is a, 
prove that the coefficient of friction is n = tan a. 

3.55. A 400 kg mass on a 30° inclined plane is acted upon by a force of 4800 newtons at angle 30° 
with the incline, as shown in Fig. 3-20. Find the acceleration of the mass if the incline (a) is 
frictionless, (6) has coefficient of friction 0.2. Ans. (a) 5.5 m/sec 2 , (6) 5.0 m/sec 2 





Fig. 3-20 



Fig. 3-21 



3.56. Work Problem 3.55 if the force of 4800 newtons acts as shown in Fig. 3-21. 
Ans. (a) 5.5 m/sec 2 , (6) 2.6 m/sec 2 

STATICS IN A UNIFORM GRAVITATIONAL FIELD 

3.57. A 100 kg weight is suspended vertically from the center of a rope as shown in Fig. 3-22. 
Determine the tension T in the rope. Ans. T = 100 kg wt = 980 nt 



y///////////////^^^^^ 






Fig. 3-23 



Fig. 3-24 



3.58. In Fig. 3-23, AB and AC are ropes attached to the ceiling CD and wall BD at C and B respectively. 
A weight W is suspended from A. If the ropes AB and AC make angles e x and e 2 with the 
wall and ceiling respectively, find the tensions T t and T 2 in the ropes. 

W cos 6 2 _ W sin e x 

T 2 — 



Ans. T, = 



cos (0x — e 2 ) ' 



cos (e 1 — e 2 ) 



3.59. Find the magnitude of the force F needed to keep mass m in equilibrium on the inclined plane 
of Fig. 3-24 if (a) the plane is smooth, (6) the plane has coefficient of friction p. 

. . . _ mg sin a ... „ __ mg (sin a — fi cos a) 
AnS ' {a) F = I^T' (6) F c^sl 



CHAP. 3] MOTION IN A UNIFORM FIELD. FALLING BODIES AND PROJECTILES 81 

3.60. How much force is needed to pull a train weighing 320 tons from rest to a speed of 15 mi/hr 
in 20 seconds if the coefficient of friction is 0.02 and (a) the track is horizontal, (6) the track is 
inclined at an angle of 10° with the horizontal and the train is going upward? [Use sin 10° = .1737, 
cos 10° = .9848.] Ans. (a) 17.4 tons, (6) 129.6 tons 

3.61. Work Problem 3.60(6) if the train is going down the incline. Ans. 3.6 tons 

3.62. A train of mass m is coasting down an inclined plane of angle a and coefficient of friction fi 
with constant speed v Q . Prove that the force needed to stop the train in a time t is given by 
Wfl^sin a — (i cos a) + mt> /r. 

MISCELLANEOUS PROBLEMS 

3.63. A stone is dropped down a well and the sound of the splash is heard after time t. A ssuming the 
speed of sound is c, prove that the depth of the water level in the well is (y/c 2 + Igcr — c) 2 /2g. 

3.64. A projectile is launched downward from the top of an inclined plane of angle a in a direction 
making an angle y with the incline. Assuming that the projectile hits the incline, prove that 

2i>o sin y cos (y — a) ,,,,., . , ., 

(a) the ranee is given by R = 5 and that (6) the maximum range down the 

v ' 9 9 cos 2 a 

K 

g(l — sin a) 



incline is i? max — ~" n — 



3.65. A cannon is located on a hill which has the shape of an inclined plane of angle a with the horizontal. 
A projectile is fired from this cannon in a direction up the hill and making an angle /3 with it. Prove 

/ 2 sin 2a * 
that in order for the projectile to hit the hill horizontally we must have = tan M 3 _ cog 2a 

3.66. Suppose that two projectiles are launched at angles a and /3 with the horizontal from the 
same place at the same time in the same vertical plane and with the same initial speed. Prove 
that during the course of the motion, the line joining the projectiles makes a constant angle 
with the vertical given by £(a + /3). 

3.67. Is it possible to solve equation (1), page 33, by the method of separation of variables? Explain. 

3.68. When launched at angle 9 1 with the horizontal a projectile falls a distance D x short of its target, 
while at angle 2 it falls a distance D 2 beyond the target. Find the angle at which the projectile 
should be launched so as to hit the target. 

3.69. An object was thrown vertically downward. During the tenth second of travel it fell twice as far 
as during the fifth second. With what speed was it thrown? Ans. 16 ft/sec 

3.70. A gun of muzzle speed v is situated at height h above a horizontal plane. Prove that the angle 
at which it must be fired so as to achieve the greatest range on the plane is given by 
e = % cos -1 gh/(vl + gh). 

3.71. In Fig. 3-25, AB is a smooth table and masses ra x 
and w 2 are connected by a string over the smooth — 
peg at B. Find (a) the acceleration of mass m 2 
and (6) the tension in the string. 

m 2 — m x 
AnS - (o) ^T+nT 2 g ' m2>mi 

jm^ Fig>3 . 25 

Wi + m 2 

3.72. Work Problem 3.71 if the table AB has coefficient of friction ft. 

3.73. The maximum range of a projectile when fired down an inclined plane is twice the maximum 
range when fired up the inclined plane. Find the angle which the incline makes with the horizontal. 
Ans. sin -1 1/3 



I I -JJ 



B 



m 2 



82 



MOTION IN A UNIFORM FIELD. FALLING BODIES AND PROJECTILES [CHAP. 3 



3.74. Masses m-^ and ra 2 are located on smooth inclined planes 
of angles a t and a 2 respectively and are connected by 
an inextensible string of negligible mass which passes 
over a smooth peg at A [Fig. 3-26]. Find the accelera- 
tions of the masses. 

Ans. The accelerations are in magnitude equal to 
Wi sin a x — m 2 sin a 2 



m x + m 2 



9 




Fig. 3-26 



3.75. Work Problem 3.74 if the coefficient of friction between the masses and the incline is ju. 



Ans. 



Wj sin <*! — m 2 sin a 2 — nm x cos a x — [im 2 cos a 2 
mi + m 2 



3.76. Prove that the least horizontal force F needed to pull 
a cylinder of radius a and weight W over an obstacle 
of height b [see Fig. 3-27] is given in magnitude by 
Wy/b(2a-b)/(a - 6). 

3.77. Explain mathematically why a projectile fired from 
cannon A at the top of a cliff at height H above the 
ground can reach a cannon B located on the ground, 
while a projectile fired from cannon B with the same 
muzzle velocity will not be able to reach cannon A. 

3.78. In Fig. 3-28 the mass m hangs from an inextensible string OA. 
It is pulled aside by a horizontal string AB so that OA makes 
an angle a with the vertical. Find the tension in each string. 
Ans. Tension in AB = mg tan a; in OA = mg sec a 

3.79. A particle moving along the x axis is acted upon by a resisting 
force which is such that the time t for it to travel a distance x 
is given by t = Ax 2 + Bx + C where A, B and C are constants. 
Prove that the magnitude of the resisting force is proportional 
to the cube of the instantaneous speed. 




Fig. 3-27 

y////////////. 

O 




3.80. 



3.81. 



3.82. 



3.83. 



3.84. 



A projectile is to be launched so as to go from A to B 
[which are respectively at the bases of a double inclined 
plane having angles a and /3 as shown in Fig. 3-29] and 
just barely miss a pole of height H. If the distance 
between A and B is D, find the angle with the horizontal 
at which the projectile should be launched. 

A particle of mass m moves on a frictionless inclined 
plane of angle a and length I. If the particle starts 
from rest at the top of the incline, what will be its 
speed at the bottom assuming that air resistance is equal 
to kv where v is the instantaneous speed and k is con- 
stant? 




Fig. 3-29 



Suppose that in Problem 3.23 the particle P is given an initial speed v Q at the top of the circle 
(or sphere), (a) Prove that if v ^ yfgb, the angle e at which the particle leaves the circle is given 
by sin -1 (§ + v^/Zgb). (b) Discuss what happens if v Q > ygb. 

A cannon is situated at the top of a vertical cliff overlooking the sea at height H above sea level. 
What should be the least muzzle velocity of the cannon in order that a projectile fired from it 
will reach a ship at distance D from the foot of the cliff? 

In Problem 3.83, (a) how long would it take the projectile to reach the ship and (6) what is the 
velocity on reaching the ship? 



CHAP. 3] MOTION IN A UNIFORM FIELD. FALLING BODIES AND PROJECTILES 



83 



3.85. A uniform chain of total length a has a portion 
< b < a hanging over the edge of a smooth table 
AB [see Fig. 3-30]. Prove that the time taken for 
the chain to slide off the table if it starts from rest 
is y/a/g\n(a + y/a 2 -b 2 )/b. 




Fig. 3-30 



3.86. If the table in Problem 3.85 has coefficient of friction p, prove that the time taken is 

In 



V 



a 



(i + fig 



a + V« 2 - [6(1 + fi~ H 2 



6(1 + n) — an 



3.87. A weight W x hangs on one side of a smooth fixed pulley of neg- 
ligible mass [see Fig. 3-31]. A man of weight W 2 pulls himself 
up so that his acceleration relative to the fixed pulley is a. 
Prove that the weight W^ moves upward with acceleration given by 

[9(W 2 -W 1 )-W 2 a]/W 1 . 

3.88. Two monkeys of equal weight are hanging from opposite ends of a 
rope which passes over a smooth fixed pulley of negligible mass. 
The first monkey starts to climb the rope at a speed of 1 ft/sec 
while the other remains at rest relative to the rope. Describe the 
motion of the second monkey. 

Ans. The second monkey moves up at the rate of 1 ft/sec. 



W////////A 




Fig. 3-31 



3.89. Prove that the particle of Problem 3.23 will land at a distance from the base of the sphere 
given by (4\^90 + 19^)6/81. 

3.90. Prove that if friction is negligible the time taken for a particle to slide down any chord of a 
vertical circle starting from rest at the top of the circle is the same regardless of the chord. 



3.91. Given line AB of Fig. 3-32 and point P where AB and 
P are in the same vertical plane. Find a point Q on 
AB such that a particle starting from point P will 
reach Q in the shortest possible time. 

[Hint. Use Problem 3.90.] 

3.92. Show how to work Problem 3.91 if line AB is re- 
placed by a plane curve. Can it be done for a space 
curve? Explain. 




3.93. Find the work done in moving the mass from the top of the incline of Problem 3.18 to the bottom. 
Ans. mgl(sm a — n cos a) 

3.94. The force on a particle having electrical charge q and which is moving in a magnetic field of intensity 
or strength B is given by F = g(v X B) where v is the instantaneous velocity. Prove that if the 
particle is given an initial speed v in a plane perpendicular to a magnetic field B of constant 
strength, then it (a) will travel with constant speed v Q and (6) will travel in a circular path 
of radius mv /qB. Assume that gravitational forces are negligible. 

3.95. Prove that the period, i.e. the time for one complete vibration, of the particle of Problem 3.94 
is independent of the speed of the particle and find its value. Ans. 2irm/qB 



3.96. Work Problem 3.94 if B is constant and the particle is given an initial speed v in a plane which 
is not necessarily perpendicular to the magnetic field. Can we define a period in this case? Explain. 



84 



MOTION IN A UNIFORM FIELD. FALLING BODIES AND PROJECTILES [CHAP. 3 



3.97. If a particle of electrical charge q and mass m moves with velocity v in an electromagnetic field 
having electric intensity E and magnetic intensity B the force acting on it, called the Lorentz force, 
is given by 

F = g(E + vXB) 

Suppose that B and E are constant and in the directions of the negative y and positive z axes 
respectively. Prove that if the particle starts from rest at the origin, then it will describe a 
cycloid in the yz plane whose equation is 

y = b(e — sin e), z = 6(1 — cos e) 

where = qBt/m, b = mE/qB 2 and t is the time. 

3.98. (a) An astronaut of 80 kg wt on the earth takes oft" vertically in a space ship which achieves 
a speed of 2000 km/hr in 2 minutes. Assuming the acceleration to be constant, what is his apparent 
weight during this time? (b) Work part (a) if the astronaut has 180 lb wt on the earth and the 
space ship achieves a speed of 1280 mi/hr in 2 minutes. Ans. (a) 117 kg wt, (6) 268 lb wt 



3.99. 



In Problem 3.82, how far from the base of the sphere will the particle land? 



3.100. In Fig. 3-33 weight W t is on top of weight W 2 which is in turn on a horizontal plane. The 
coefficient of friction between W x and W 2 is ^ while that between W 2 and the plane is fi 2 . Suppose 
that a force F inclined at angle a to the horizontal is applied to weight W^. Prove that if 
cot a ^ fii > n 2 , then a necessary and sufficient condition that W 2 move relative to the plane 
while Wi not move relative to W 2 is that 



COS a — n 2 sin a 



< F g 



COS a — jii Sin a 




Fig. 3-33 



3.101. Discuss the results in Problem 3.100 if any of the conditions are not satisfied. 

3.102. Give a generalization of Problem 3.100. 

3.103. Describe the motion of the particle of Problem 3.97 if E and B are constants, and have the same 
direction. 

3.104. A bead of mass m is located on a parabolic wire with its axis 
vertical and vertex directed downward as in Fig. 3-34 and 
whose equation is cz = x 2 . If the coefficient of friction is 
fi, find the highest distance above the x axis at which the 
particle will be in equilibrium. Ans. \p?c 

3.105. Work Problem 3.104 if the parabola is replaced by a vertical 
circle of radius 6 which is tangent to the x axis. 




Fig. 3-34 



3.106. A weight W is suspended from 3 equal strings of length I which are attached to the 3 vertices 
of a horizontal equilateral triangle of side s. Find the tensions in the strings. 

Ans. Wl/^/91 2 - 3s 2 

3.107. Work Problem 3.106 if there are n equal strings attached to the n vertices of a regular polygon 
having n sides. 



CHAP. 3] MOTION IN A UNIFORM FIELD. FALLING BODIES AND PROJECTILES 



85 



3.108. 



3.109. 



A rope passes over a fixed pulley A of Fig. 3-35. At one end 
of this rope a mass M l is attached. At the other end of the 
rope there is a pulley of mass M 2 over which passes another 
rope with masses m 1 and ra 2 attached. Prove that the accel- 
eration of the mass m t is given by 

3m 2 M 2 — m^M x — m 1 M 2 — m 2 M l — Am 1 m 2 



(m x + m^Mi + M 2 ) + 4m x m 2 



9 



An automobile of weight W with an engine having constant 
instantaneous power "P, travels up an incline of angle a. 
Assuming that resistance forces are r per unit weight, prove 
that the maximum speed which can be maintained up the 



incline is 



W(r + sin a) 



S7\ 

KM 



d 






Fig. 3-35 



3.110. An automobile of weight W moves up an incline of angle a, powered by an engine having 
constant instantaneous power e P. Assuming that the resistance to motion is equal to kv per unit 
weight where v is the instantaneous speed and k is a constant, prove that the maximum speed 
which is possible on the incline is (y/W 2 sin 2 a + 4k WP — W sin <x)/2kW. 



3.111. A chain hangs over a smooth peg with length a on one side and length 6, where < b < a, on the 

U+b /v / a+V / & > 
other side. Prove that the time taken for the chain to slide off is given by \ j — - — In — — — 



3.112. Prove that a bead P which is placed anywhere on a vertical frictionless wire [see Fig. 3-36] in the 
form of a cycloid 

x — b(e + sin e), y = 6(1 — cos e) 

will reach the bottom in the same time regardless of the starting point and find this time. 
Ans. vy/b/g 

y 




Fig. 3-36 



Chapter J,, 



The SIMPLE HARMONIC 

OSCILLATOR and the 

SIMPLE PENDULUM 



E 



-nm^ 



m 







(a) 



l + x 



THE SIMPLE HARMONIC OSCILLATOR 

In Fig. 4-l(a) the mass m lies on a friction- 
less horizontal table indicated by the x axis. 
It is attached to one end of a spring of negligible 
mass and unstretched length I whose other end 
is fixed at E. 

If m is given a displacement along the x axis 
[see Fig. 4-1(6)] and released, it will vibrate or 
oscillate back and forth about the equilibrium 
position 0. 

To determine the equation of motion, note 
that at any instant when the spring has length 
I + x [Fig. 4-1(6)] there is a force tending to re- 
store m to its equilibrium position. According 
to Hooke's law this force, called the restoring 
force, is proportional to the stretch x and is 
given by 

F R = - K xi (1) 

where the subscript R stands for "restoring force" and where « is the constant of propor- 
tionality often called the spring constant, elastic constant, stiffness factor or modulus of 
elasticity and i is the unit vector in the positive x direction. By Newton's second law we have 



E 



-nmr^ 




Fig. 4-1 



d 2 (xi) 
m ~aW = ~ KX1 



or mx + kx 



(2) 



This vibrating system is called a simple harmonic oscillator or linear harmonic oscillator. 
This type of motion is often called simple harmonic motion. 



AMPLITUDE, PERIOD AND FREQUENCY 
OF SIMPLE HARMONIC MOTION 

If we solve the differential equation (2) subject to the initial conditions x = A and 
dx/dt = at t = 0, we find that 



(3) 



x = A cos o>t where o> = s/K/m 

For the case where A = 20, m = 2 and * = 8, see Problem 4.1. 

Since cos tat varies between —1 and +1, the mass oscillates between x = —A and x = A. 
A graph of x vs. t appears in Fig. 4-2. 



86 



CHAP. 4] THE SIMPLE HARMONIC OSCILLATOR AND THE SIMPLE PENDULUM 87 




Fig. 4-2 

The amplitude of the motion is the distance A and is the greatest distance from the 
equilibrium position. 

The period of the motion is the time for one complete oscillation or vibration [some- 
times called a cycle] such as, for example, from x = A to x = —A and then back to 
x = A again. If P denotes the period, then 

p = 2ttU = 2ir^/mU {U) 

The frequency of the motion, denoted by /, is the number of complete oscillations or 
cycles per unit time. We have 

f = P = 2^ = 2^\m (5) 

In the general case, the solution of (2) is 

x = A cos mt + B sin o>t where w = VkTm (6) 

where A and B are determined from initial conditions. As seen in Problem 4.2, we can 
write (6) in the form 

x = C cos (<at — <£) where <o = y/T/m (7) 

and where C = y/A 2 + B 2 and <f> = tan' 1 (B/A) (8) 

The amplitude in this case is C while the period and frequency remain the same as in 
(4) and (5), i.e. they are unaffected by change of initial conditions. The angle <f> is called 
the phase angle or epoch chosen so that ^ </> ^ v. If = 0, (7) reduces to (3). 

ENERGY OF A SIMPLE HARMONIC OSCILLATOR 

If T is the kinetic energy, V the potential energy and E = T + V the total energy of a 
simple harmonic oscillator, then we have 

T = %mv 2 , V = $kx 2 (9) 

and E = \mv 2 + \ K x 2 (10) 

See Problem 4.17. 

THE DAMPED HARMONIC OSCILLATOR 

In practice various forces may act on a harmonic oscillator, tending to reduce the 
magnitude of successive oscillations about the equilibrium position. Such forces are some- 
times called damping forces. A useful approximate damping force is one which is propor- 
tional to the velocity and is given by 

dx 
F D = -Pv = -Pvi = -jB-^i (11) 



88 



THE SIMPLE HARMONIC OSCILLATOR AND THE SIMPLE PENDULUM [CHAP. 4 



where the subscript D stands for "damping force" and where /? is a positive constant 
called the damping coefficient. Note that F D and v are in opposite directions. 

If in addition to the restoring force we assume the damping force (11), the equation 
of motion of the harmonic oscillator, now called a damped harmonic oscillator, is given by 



d 2 x n dx d 2 x , n dx , A 

m-T7z — —kx — /? — or m^ns + /3~rr + kX - 



dt 2 ~ *" M dt WA "" dt 2 ' ^ dt 
on applying Newton's second law. Dividing by m and calling 

/?/m = 2y, kItti = <o 2 
this equation can be written 

X + 2y# + o> 2 X = 

where the dots denote, as usual, differentiation with respect to t. 



(12) 

(13) 
(U) 



OVER-DAMPED, CRITICALLY DAMPED AND 
UNDER-DAMPED MOTION 

Three cases arise in obtaining solutions to the differential equation (14). 

Case 1, Over-damped motion, y 2 > <o 2 , i.e. /? 2 > 4«ra 
In this case (14) has the general solution 

x = e' yt (Ae at + Be~ at ) where a = yV - «> 2 i 15 ) 

and where the arbitrary constants A and B can be found from the initial conditions. 

Case 2, Critically damped motion, y 2 '= w 2 , i.e. p 2 = 4 K m 
In this case (14) has the general solution 

x = e~ yt (A + Bt) (16) 

where A and B are found from initial conditions. 

Case 3, Under-damped or damped oscillatory motion, y 2 < <o 2 , i.e. /3 2 < 4*<ra 

In this case (14) has the general solution 

x = e~ yt (A sin \t + B cos \t) 

= Ce~ yt cos (\t-<t>) where A = V«> 2 - Y 2 i 17 ) 

and where C = V^ 2 + ^ 2 » called the amplitude and <£, called the phase angle or epoch, 
are determined from the initial conditions. 

In Cases 1 and 2 damping is so large that no 
oscillation takes place and the mass m simply 
returns gradually to the equilibrium position 
x = 0. This is indicated in Fig. 4-3 where we 
have assumed the initial conditions x = x , 
dx/dt — 0. Note that in the critically damped 
case, mass m returns to the equilibrium position 
faster than in the over-damped case. 

In Case 3, damping has been reduced to such 
an extent that oscillations about the equilibrium 
position do take place, although the magnitude 
of these oscillations tend to decrease with time 
as indicated in Fig. 4-3. The difference in times 



Critically damped motion, y* = a? 

Over-damped motion, y* > «* 




Under-damped motion, y 2 < u* 
Fig. 4-3 



CHAP. 4] THE SIMPLE HARMONIC OSCILLATOR AND THE SIMPLE PENDULUM 89 

between two successive maxima [or minima] in the under-damped [or damped oscillatory] 
motion of Fig. 4-3 is called the period of the motion and is given by 

p = ?![ = 2tt _ 47rm , lg . 

A \A> 2 - y 2 \/4kWI - p 2 

and the frequency, which is the reciprocal of the period, is given by 

1 _ k_ V^ 7 ?" _ V^m - p 2 
f ~ P ~ 2tt ~ 2tt ~ 47rm y ' 

Note that if p - 0, (18) and (19) reduce to (4) and (5) respectively. The period and 
frequency corresponding to /? = are sometimes called the natural period and natural 
frequency respectively. 

The period P given by (18) is also equal to two successive values of t for which 
cos(A£ — <£) = 1 [or cos(Ai — <j>) = —1] as given in equation (17). Suppose that the values 
of x corresponding to the two successive values t n and t n +i = t n + P are x n and x n +i respec- 
tively. Then 

xjxn+i = e-^/e-^ +P) = e^ p (20) 

The quantity 8 = In (xjx n +i) = yP (21) 

which is a constant, is called the logarithmic decrement. 



FORCED VIBRATIONS 

Suppose that in addition to the restoring force — kx\ and damping force —fSvi we impress 
on the mass m a force F(t)i where 

F(t) = F cos at (22) 

Then the differential equation of motion is 

d 2 x dx 

m -rp = —kX - /? -7T + Fq cos at (23) 

Or X + 2yi + oy 2 X = f Q COS at (2U) 

where y = p/2m, w 2 = K /m, f = FJm (25) 

The general solution of (24) is found by adding the general solution of 

X + 2yX + o> 2 X = (26) 

[which has already been found and is given by (15), (16) or (17)] to any particular solution of 
(24). A particular solution of (24) is given by [see Problem 4.18] 

x = ° cos (at - «/,) (27) 

vV-<o 2 ) 2 + 4yV 

where tan d, = * ya . ^ «/» ^ tt (^5) 

a — w 

Now, as we have seen, the general solution of (26) approaches zero within a short time 
and we thus call this solution the transient solution. After this time has elapsed, the motion 
of the mass m is essentially given by (27) which is often called the steady-state solution. 
The vibrations or oscillations which take place, often called forced vibrations or forced 
oscillations, have a frequency which is equal to the frequency of the impressed force but 
lag behind by the phase angle 4>. 



90 



THE SIMPLE HARMONIC OSCILLATOR AND THE SIMPLE PENDULUM [CHAP. 4 



RESONANCE 

The amplitude of the steady-state oscillation (27) is given by 

/ 
oA = ° = (29) 

VV-a> 2 ) 2 + 4 y V 

assuming y ^ 0, i.e. p ¥° 0, so that damping is assumed to be present. The maximum value 
of cA in this case occurs where the frequency «/2tt of the impressed force is such that 

a 2 = a 2 R = w 2 -2y 2 (30) 

assuming that y 2 < |« 2 [see Problem 4.19]. Near this frequency very large oscillations may 
occur, sometimes causing damage to the system. The phenomenon is called resonance and 
the frequency ajl-n is called the frequency of resonance or resonant frequency. 

The value of the maximum amplitude at the resonant frequency is 

u 



cA, 



2y\A, 2 - y 2 

The amplitude (29) can be written in terms of a R as 



aA = 



VV-a 2 ) 2 + 4yV-y 2 ) 



(31) 



(32) 



A graph of cA vs. a 2 is shown in Fig. 4-4. Note that the graph is symmetric around the 
resonant frequency and that the resonant frequency, frequency with damping and natural 
frequency (without damping) are all different. In case there is no damping, i.e. y = or 
/? = 0, all of these frequencies are identical. In such case resonance occurs where the 
frequency of the impressed force equals the natural frequency of oscillation. The general 
solution for this case is 

x = A cos <at + B sin <at + 75— sin at (33) 

From the last term in (33) it is seen that the oscillations build up with time until finally 
the spring breaks. See Problem 4.20. 



qA 



Resonant frequency 

Frequency with damping 



Natural frequency 
without damping) 




v/////////////// ( 



Fig. 4-4 




Fig. 4-5 



THE SIMPLE PENDULUM 

A simple pendulum consists of a mass m [Fig. 4-5] at the end of a massless string or rod 
of length I [which always remains straight, i.e. rigid]. If the mass m, sometimes called the 
pendulum bob, is pulled aside and released, the resulting motion will be oscillatory. 

Calling the instantaneous angle which the string makes with the vertical, the 
differential equation of motion is [see Problem 4.23] 



CHAP. 4] THE SIMPLE HARMONIC OSCILLATOR AND THE SIMPLE PENDULUM 91 

assuming no damping forces or other external forces are present. 

For small angles [e.g. less than 5° with the vertical], sin 9 is very nearly equal to 0, where 
9 is in radians, and equation (3U) becomes, to a high degree of approximation, 

§ = -f» w 

This equation has the general solution 

9 = A cos Vgllt + Bsin^/gllt (36) 

where A and B are determined from initial conditions. For example, if 9 = 0o, = at 

t = 0, then , 

9 = 6o cos Vgllt (37) 

In such case, the motion of the pendulum bob is that of simple harmonic motion. The period 

is given by , 

P = 2-xyJTig (38) 

and the frequency is given by 

f = ¥ = ^^ fl {39) 

If the angles are not necessarily small, we can show [see Problems 4.29 and 4.30] 
that the period is equal to 

iir/2 dB 



P = 4 



fir. 



Fj = — Kjzi 



VI - W sin 2 9 

where k = sin (9 /2). For small angles this reduces to (38). 

For cases where damping and other external forces are considered, see Problems 4.25 
and 4.114. 

THE TWO AND THREE DIMENSIONAL HARMONIC OSCILLATOR 

Suppose a particle of mass m moves in the xy plane y 

under the influence of a force field F given by 

F = - Kl a?i - K 2 yj (U) 

where k x and k 2 are positive constants. 

In this case the equations of motion of m are 
given by j 

m dP = ~ K i x ' m d& = ~ K * y ^ 

and have solutions Fig. 4-6 

x = A 1 cos -y/ajm t + B x sin yj«.jm t, y = A 2 cos yj*.jm t + B 2 sin y/ajm t (A3) 

where Ai, B\, A%, B 2 are constants to be determined from the initial conditions. The mass m 
subjected to the force field (41) is often called a two-dimensional harmonic oscillator. The 
various curves which m describes in its motion are often called Lissajous curves or figures. 

These ideas are easily extended to a three dimensional harmonic oscillator of mass m 
which is subject to a force field given by 

F = -k x x\ - K 2 y] - K 3 zk (U) 

where < lf k 2 , « 3 are positive constants. 



m 

F 2 = -/c 2 l/j 



92 THE SIMPLE HARMONIC OSCILLATOR AND THE SIMPLE PENDULUM [CHAP. 4 



Solved Problems 

SIMPLE HARMONIC MOTION AND THE 
SIMPLE HARMONIC OSCILLATOR 

4.1. A particle P of mass 2 moves along the x axis attracted toward origin O by a force 
whose magnitude is numerically equal to Sx [see Fig. 4-7]. If it is initially at rest 
at x = 20, find (a) the differential equation and initial conditions describing the 
motion, (b) the position of the particle at any time, 
(c) the speed and velocity of the particle at any time, 
and (d) the amplitude, period and frequency of the 
vibration. 



-8a?i 



xi 



(a) Let r = xi be the position vector of P. The acceleration { 

d 2 d 2 x 

of P is jijjC^i) ~ ~d+2*' ^ ne ne t f° rce acting on P is 

— 8xi. Then by Newton's second law, Fig. 4-7 

n d 2 x. „ . d 2 x , . .„. 

2-^1 = -8jbi or ^p-+ 4x = (1) 

which is the required differential equation of motion. The initial conditions are 

x = 20, dx/dt = at t = (2) 

(6) The general solution of (1) is 

x = A cos It + B sin 2£ (3) 

When t = 0, x = 20 so that A = 20. Thus 

x = 20 cos It + B sin 2t (4) 

Then dx/dt = -40 sin 2t + 2B cos 2t (5) 

so that on putting t = 0, dx/dt = we find B = 0. Thus (5) becomes 

a; = 20 cos 2i (0) 

which gives the position at any time. 

(c) From (6) dx/dt = -40 sin It which gives the speed at any time. The velocity is given by 

^i = -40sin2ti 
dt 

(d) Amplitude = 20. Period = 2tt/2 = w. Frequency = 1/period = 1/v. 



4.2. (a) Show that the function A cos at + B sin «>t can be written as C cos (U - </>) 
where C = V A 2 + B 2 and <j> = tan -1 (S/A). (6) Find the amplitude, period and 
frequency of the function in (a). 

(a) A cos at + B sin at = \l A 2 + B 2 ( cos at + sin w A 

W^ 2 + B 2 VA 2 + B 2 / 

= \/A 2 + # 2 (cos <(> cos w£ + sin <f> sin to*) 



= VA 2 + B 2 cos («t - *) = Ccos(wt-0) 



where cos = A/ VA 2 + £ 2 and sin = B/y/A 2 + B 2 , i.e. tan <p = 5/ A or = tan- 1 £/A, 
and C = VA 2 + £ 2 . We generally choose that value of <p which lies between 0° and 180°, 
i.e. g <p ^ v. 

(b) Amplitude = maximum value = C - y/A^TB 2 . Period = 2a-/«. Frequency = w/2*-. 



CHAP. 4] THE SIMPLE HARMONIC OSCILLATOR AND THE SIMPLE PENDULUM 93 

4.3. Work Problem 4.1 if P is initially at x = 20 but is moving (a) to the right with speed 
30, (b) to the left with speed 30. Find the amplitude, period and frequency in 
each case. 

(a) The only difference here is that the condition dx/dt = at * = of Problem 4.1 is replaced 
by dx/dt = 30 at t ~ 0. Then from (5) of Problem 4.1 we find B = 15, and (3) of 

Problem 4.1 becomes 

x = 20 cos 2t + 15 sin 2t CO 



\ 



which gives the position of P at any time. This may be written [see Problem 4.2] as 

x = V(20) 2 + (15)2 I 20 — cos 2t + 15 ; sin 2t \ 

1/(20)2+ (15) 2 V(20)2 + (15)2 J 

= 25{|cos2i + f sin2£} = 25 cos (2t - 0) 
where cos = f, sin = f (2) 

The angle which can be found from (2) is often called the phase angle or epoch. 

Since the cosine varies between —1 and +1, the amplitude = 25. The period and fre- 
quency are the same as before, i.e. period = 2w/2 = w and frequency = 2/2?r = II v. 

(b) In this case the condition dx/dt = at t = of Problem 4.1 is replaced by dx/dt = -30 
at t = 0. Then B — —15 and the position is given by 

x = 20 cos It — 15 sin 2t 
which as in part (a) can be written 

x = 25{|cos2t- f sin2t} 

= 25{cos cos 2t + sin ^ sin 2i) = 25 cos (2t — 0) 
where cos f = f , sin ^ = — f . 

The amplitude, period and frequency are the same as in part (a). The only difference 
is in the phase angle. The relationship between ^ and <f> is $ = <p + w. We often describe 
this by saying that the two motions are 180° out of phase with each other. 



4.4. A spring of negligible mass, suspended vertically from one end, is stretched a 
distance of 20 cm when a 5 gm mass is attached to the other end. The spring and 
mass are placed on a horizontal frictionless table as in Fig. 4-l(a), page 86, with the 
suspension point fixed at E. The mass is pulled away a distance 20 cm beyond the 
equilibrium position O and released. Find (a) the differential equation and initial 
conditions describing the motion, (b) the position at any time t, and (c) the amplitude, 
period and frequency of the vibrations. 

(a) The gravitational force on a 5 gm mass [i.e. the weight of a 5 gm mass] is 5g = 5(980) dynes = 
4900 dynes. Then since 4900 dynes stretches the spring 20 cm, the spring constant is 
K = 4900/20 = 245 dynes/cm. Thus when the spring is stretched a distance x cm beyond the 
equilibrium position, the restoring force is — 245xi. Then by Newton's second law we have, 
if r = xi is the position vector of the mass, 

5 *g*> = _245zi or ^+ 49* = (1) 

The initial conditions are x = 20, dx/dt = at t = (2) 

(b) The general solution of (1) is x = A cos It + B sin It ($) 
Using the conditions (*) we find A = 20, B = so that x - 20 cos It. 

(c) From x = 20 cos It we see that: amplitude = 20 cm; period = 2tt/7 sec; frequency = 7/2ir 
vib/sec or l/2v cycles/sec. 



94 



THE SIMPLE HARMONIC OSCILLATOR AND THE SIMPLE PENDULUM [CHAP. 4 



4.5. A particle of mass m moves along the x axis, attracted toward a fixed point O 
on it by a force proportional to the distance from O. Initially the particle is at 
distance x from and is given a velocity v away from 0. Determine (a) the 
position at any time, (b) the velocity at any time, and (c) the amplitude, period, 
frequency, and maximum speed. 

(a) The force of attraction toward O is — kx'i where k is a 
positive constant of proportionality. Then by Newton's 
second law, 



d?x. 



•• . KX 

—kxi or x H =0 

m 



Solving (1), we find 

x = A cos y/ic/m t + B sin yjn/m t 
We also have the initial conditions 



(1) 
(2) 





F = — Kxi 









m 



Fig. 4-8 



x = x , dx/dt — v at t = 
From x = x at t = we find, using (2), that A = x . Thus 

x = x cos V ' ii/m t + B sin yic/m t 
so that dx/dt = —x Q yjulm sin V*/™ t + B y/lc/m cos yV™ £ 

From dx/dt = v at t = we find, using (5), that B = v vWk. Thus (4) becomes 

a; = #0 cos V~i</ni t + v Q \Zm/K sin vk/w t 
Using Problem 4.2, this can be written 



where 



x = yx^+mvUic cos (vk/ot £ — 0) 
^ = tan -1 (v /x ) yW/c 



(3) 

(•4) 
(5) 

(8) 

(7) 
(8) 



(6) The velocity is, using (0) or (7), 

dx . 



v = -tt i = (— #o y/x/m sin VxTm £ + v cos yVm £) i 
= —^Jit/m \/x 2 + mv 2 /ic sin (yfic/mt — <j>) i 
= — V^n + k«?/w. sin {\[7hn t — 0) i 



(3) 



(c) The amplitude is given from (7) by yxJ+mv*/K. 

From (7), the period is P = 2srV 'ic/m. The frequency is / = 1/P = 2iry/m/K. 



From (3), the speed is a maximum when sin (y/T/mt -</>) = ±1; this speed is v^ 2 , + kb 2 /™. 



4.6. An object of mass 20 kg moves with simple harmonic motion on the x axis. Initially 
(t = 0) it is located at the distance 4 meters away from the origin x = 0, and has 
velocity 15 m/sec and acceleration 100 m/sec 2 directed toward * = 0. Find (a) the 
position at any time, (b) the amplitude, period and frequency of the oscillations, and 
(c) the force on the object when t - -rr/10 sec. 



(a) If x denotes the position of the object at time t, then the initial conditions are 

x = 4, dx/dt = -15, d 2 x/dt 2 = -100 at t = 



Now for simple harmonic motion, 



= A cos u>t + B sin at 



Differentiating, we find 



dx/dt = —Aw sin at + Ba cos at 
d 2 x/dt 2 = —Aw 2 cos ut - Ba 2 sin at 



(1) 



(3) 
(4) 



CHAP. 4] THE SIMPLE HARMONIC OSCILLATOR AND THE SIMPLE PENDULUM 



95 



Using conditions (1) in {2), (S) and (4), we find 4 = A, -15 = Bw, -100 = -Aw 2 . Solving 
simultaneously, we find A = 4, w = 5, 5 = — 3 so that 



x = 4 cos 5t — 3 sin 5t 
which can be written 

x = 5 cos (5t — #) where cos ^ = ^ , sin ^ = — § 

(6) From (tf) we see that: amplitude = 5 m, period = 2jt/5 sec, frequency = 5/2jr vib/sec. 

(c) Magnitude of acceleration = d 2 x/dt 2 = —100 cos 5t + 75 sin 5* = 75 m/sec 2 at t = tt/10. 
Force on object = (mass) (acceleration) = (20 kg)(75 m/sec 2 ) = 1500 newtons. 



(5) 
(6) 



4.7. 



4.8. 



A 20 lbwt object suspended from the end of a ver- 
tical spring of negligible mass stretches it 6 inches. 
(a) Determine the position of the object at any time 
if initially it is pulled down 2 inches and then re- 
leased, (b) Find the amplitude, period and fre- 
quency of the motion. 

(a) Let D and E [Fig. 4-9] represent the position of the 
end of the spring before and after the object is put on 
the spring. Position E is the equilibrium position of 
the object. 

Choose a coordinate system as shown in Fig. 4-9 
so that the positive z axis is downward with origin at 
the equilibrium position. 

By Hooke's law, since 20 lb wt stretches the spring 
\ ft, 40 lb wt stretches it 1 ft; then 40(.5 + z) lb wt 
stretches it (.5 + z) ft. Thus when the object is at 
position F there is an upward force acting on it of 
magnitude 40(.5 + z) and a downward force due to its 
weight of magnitude 20. By Newton's second law we 
thus have 

H^k = 20k - 40(.5 + z)k or 



mm^x? 



D 



E 



J .5 ft 



zft 



F 



F 1 



I I 



Fig. 4-9 



W2 + 64, = 



Solving, z = A cos St + B sin St 

Now at t = 0, z = % and dz/dt = 0; thus A = £, B = and 



(1) 



z = ^ cos 8£ 



(b) From (2): amplitude = £ ft, period = 2jt/8 = v/4 sec, frequency = 4/v vib/sec. 

Work Problem 4.7 if initially the object is pulled down 3 inches (instead of 2 inches) 
and then given an initial velocity of 2 ft/sec downward. 

In this case the solution (1) of Problem 4.7 still holds but the initial conditions are: at t = 0, 
z — \ and dz/dt = 2. From these we find A = \ and B = \, so that 

z = I cos St + I sin Bt = y/2/4 cos (St - ir/4) 

Thus amplitude = \/2/4 ft, period = 2a-/8 = jt/4 sec, frequency = 4Ar vib/sec. Note that the period 
and frequency are unaffected by changing the initial conditions. 



4.9. A particle travels with uniform angular speed a around a circle of radius 6. Prove 
that its projection on a diameter oscillates with simple harmonic motion of period 
2tt/g> about the center. 

Choose the circle in the xy plane with center at the origin O as in Fig. 4-10 below. Let Q be 
the projection of particle P on diameter AB chosen along the x axis. 



96 



THE SIMPLE HARMONIC OSCILLATOR AND THE SIMPLE PENDULUM [CHAP. 4 



If the particle is initially at B, then in time t we will 
have I BOP — e — ut. Then the position of P at time t is 



r = b cos ut i + b sin ut j 
The projection Q of P on the x axis is at distance 
r • i — x = b cos ut 



(1) 



(2) 



from O at any time t. From (2) we see that the projection Q 
oscillates with simple harmonic motion of period 2ir/w about 
the center O. 




Fig. 4-10 



THE DAMPED HARMONIC OSCILLATOR 

4.10. Suppose that in Problem 4.1 the particle P has also a damping force whose magnitude 
is numerically equal to 8 times the instantaneous speed. Find (a) the position and 
(b) the velocity of the particle at any time, (c) Illustrate graphically the position of 
the particle as a function of time t. y 



(a) 



(b) 



(c) 



In this case the net force acting on P is [see 

doc 

— 8a;i — 8-rri. Then by Newton's sec- 



Fig. 4-11] 
ond law, 



2— i 



dt 



= -8«i 



d 2 x 



+ ^ + i * 



, dx 
di 















- 8^i 





















1 


Fig. 4-11 


P 





This has the solution [see Appendix, page 352, Problem C.14] 

x = e-»(A + Bt) 

When t = 0, x - 20 and dx/dt = 0; thus A = 20, B = 40, and a; = 20e~ 2t (1 + 2t) gives 
the position at any time t. 



The velocity is given by 

dx . 
V = ^ 



-80«e-«i 



The graph of a; vs. £ is shown in Fig. 4-12. It is 
seen that the motion is non-oscillatory. The par- 
ticle approaches O slowly but never reaches it. 
This is an example where the motion is critically 
damped. 




Fig. 4-12 



4.11. A particle of mass 5 gm moves along the x axis under the influence of two forces: 
(i) a force of attraction to origin O which in dynes is numerically equal to 40 times 
the instantaneous distance from O, and (ii) a damping force proportional to the 
instantaneous speed such that when the speed is 10 cm/sec the damping force is 
200 dynes. Assuming that the particle starts from rest at a distance 20 cm from O, 
(a) set up the differential equation and conditions describing the motion, (b) find 
the position of the particle at any time, (c) determine the amplitude, period and 
frequency of the damped oscillations, and (d) graph the motion. 

(a) Let the position vector of the particle P be denoted by 
r = xi as indicated in Fig. 4-13. Then the force of attrac- 
tion (directed toward O) is 

-40»i (1) 



The magnitude of the damping force / is proportional to 
the speed, so that / = (S dx/dt where /? is constant. Then 
since / = 200 when dx/dt = 10, we have /? = 20 and 
/ = 20 dx/dt. To get £, note that when dx/dt > and 
x > the particle is on the positive x axis and moving to 



O 



-20(dx/dt)i 



— 40*i 



Fig. 4-13 



CHAP. 41 THE SIMPLE HARMONIC OSCILLATOR AND THE SIMPLE PENDULUM 



97 



the right. Thus the resistance force must be directed toward the left. This can only be accom 
plished if ^ 



f = - 20 ^ 



(2) 



This same form for f is easily shown to be correct if * > 0, dx/dt < 0, x < 0, dx/dt > 0, 
x < 0, dx/dt < [see Problem 4.45]. 

Hence by Newton's second law we have 



d*x. 
b dP 1 



-20% - 40*1 
dt 



d z x , . dx . 
1M + *M + SX 







(3) 
04) 

(5) 



Since the particle starts from rest at 20 cm from O, we have 

* = 20, dx/dt = at * = 

where we have assumed that the particle starts on the positive side of the x axis [we could 
just as well assume that the particle starts on the negative side, in which case x — —20]. 



(b) x = e at is a solution of (-4) if 

a 2 + 4a + 8 = 
Then the general solution is 



i(-4 ± V16 - 32) = -2 ± 2% 



x = e~ 2t (A cos2t + B sin2t) (6) 

Since x = 20 at t = 0, we find from (6) that A = 20, i.e., 

x = e~ 2t (20 cos 2t + B sin 2t) (7) 

Thus by differentiation, 

dx/dt = (e- 2 t)(-40 sin 2* + 2B cos 2t) + (-2e~2t)(20 cos 2t + B sin 2t) (8) 

Since dx/dt = at t = 0, we have from (8), B = 20. Thus from (7) we obtain 

a; = 20e-2'(cos 2t + sin 2t) = 20\/2 e" 2 * cos (2t - ir/4) (9) 

using Problem 4.2. 

(c) From (9): amplitude = 20V2 e _2t cm, period = 2jt/2 = tt sec, frequency = 1/V vib/sec. 

(d) The graph is shown in Fig. 4-14. Note that the amplitudes of the oscillation decrease toward 
zero as t increases. 



20V2 




Fig. 4-14 



4.12. Find the logarithmic decrement in Problem 4-11. 

Method 1. The maxima (or minima) of * occur where dx/dt = 0. From (9) of Problem 4.11, 

dx/dt - -80e- 2t sin2£ = 

when t — 0, ir/2, w, 3n-/2, 2ir, 5ir/2, .... The maxima occur when t = 0, v, 2v, . . . ; the minima 
occur when t = n-/2, 3s-/2, 5xr/2, .... The ratio of two successive maxima is e -2 (0)/ e -2(7r) or 
e -2(ir)/ e -2(27r) > etc., i.e. e 2v . Then the logarithmic decrement is 8 =ln(e 2w ) = 2w. 



98 THE SIMPLE HARMONIC OSCILLATOR AND THE SIMPLE PENDULUM [CHAP. 4 

Method 2. 

From (9) of Problem 4.11, the difference between two successive values of t, denoted by t n 
and t n + 1 , for which cos(2t — tt/4) — 1 (or —1) is w, which is the period. Then 

Xn 20\/2e-2tn 

— — = — = e 27r and 8 — \n(x n /x n + 1 ) = 2b- 

x n + i 20y/2e- 2t n+i 

Method 3. From (13), (18) and (21), pages 88 and 89, we have 



8 = yP 



— ( £ \( ^ irm \ = 2 ^ft 



Then since ra = 5, p — 20, k = 40 [Problem 4.11, equation (3)], 8 = 2w. 

4.13. Determine the natural period and frequency of the particle of Problem 4.11. 

The natural period is the period when there is no damping. In such case the motion is 
given by removing the term involving dx/dt in equation (3) or (4) of Problem 4.11. Thus 

d 2 x/dt 2 + Sx = or a; = A cos 2y[2t + B sin 2y/2t 
Then: natural period = 2irl2y[2 sec = irly[2 sec; natural frequency = y[2h vib/sec. 

4.14. For what range of values of the damping constant in Problem 4.11 will the motion 
be (a) overdamped, (b) underdamped or damped oscillatory, (c) critically damped? 

Denoting the damping constant by p, equation (3) of Problem 4.11 is replaced by 

p d 2 x. n dx . .. . d 2 x , P dx _ _ A 

5 dtff = -** 1 - 40 * 1 ° r d# + 5Tt +8x ' ° 

Then the motion is: 

(a) Overdamped if (/3/5) 2 > 32, i.e. p > 20 v / 2. 

(b) Underdamped if (p/5) 2 < 32, i.e. p < 20^2. 

[Note that this is the case for Problem 4.11 where p = 20.] 

(c) Critically damped if (p/5) 2 = 32, i.e. p = 20V2. 

4.15. Solve Problem 4.7 taking into account an external damping force given numerically 
in lb wt by pv where v is the instantaneous speed in ft/sec and (a) /? = 8, (b) p = 10, 
(c) p = 12.5. 

The equation of motion is 

20 d 2 z „ dz. d 2 z , SB dz , _. . 

I^ k = 20k-40(.5 + ,)k-/ 3 |k or ^ + f ^ + 64, = 

(a) If /8 = 8, then d 2 z/dt 2 + 12.8 dz/dt + 642 = 0. The solution is 

z = e~ 6At (A cos4.8£ + J? sin 4.8t) 
Using the conditions z = 1/6, dz/dt = at * = 0, we find A = 1/6, J3 = 2/9 so that 
2 = -?-e-6.4t(3 C0S 4.8£ + 4 sin 4.8*) = Jg e ~ 8 ' 4t cos < 4 - 8t ~ 53 ° 8 '> 
The motion is damped oscillatory with period 2?r/4.8 = 5V/12 sec. 

(6) If p = 10, then d 2 z/dt 2 + Udz/dt + 64z = 0. The solution is 

z = e~ 4t (A+Bt) 
Solving subject to the initial conditions gives A = %, B = f ; then % - \e~ 4t (1 + U). 



CHAP. 4] THE SIMPLE HARMONIC OSCILLATOR AND THE SIMPLE PENDULUM 



99 



The motion is critically damped since any decrease in /? would produce oscillatory motion. 

(c) If /? = 12.5 then d 2 z/dt 2 + 20dz/dt + 64z = 0. The solution is 

z = Ae-u + Be-™ 
Solving subject to initial conditions gives A = 1/6, B = -1/24; then z - \e~^ - ^e~ 16t . 
The motion is overdamped. 



ENERGY OF A SIMPLE HARMONIC OSCILLATOR 

4.16. (a) Prove that the force F = - K xi acting on a simple harmonic oscillator is con- 
servative, (b) Find the potential energy of a simple harmonic oscillator. 



(a) We have V X F 



i J k 

d/dx d/dy d/dz 
—kx 



= so that F is conservative. 



(6) The potential or potential energy is given by V where F--VV or 

fdV. ,BV. ,8V. 

~ KXl = -(jr + ^ J + *T k 

Then dV/dx = kx, dV/dy = 0, dV/dz = from which V = fax 2 + c. Assuming V = cor- 
responding to x = 0, we find c = so that V = fax 2 . 



4.17. Express in symbols the principle of conservation of energy for a simple harmonic 
oscillator. 

By Problem 4.16(6), we have 

Kinetic energy + Potential energy = Total energy 
or fafiv 2 + fax 2 = E 

which can also be written, since v = dx/dt, as fan{dx/dt) 2 + fax 2 = E. 

Another method. The differential equation for the motion of a simple harmonic oscillator is 

md 2 x/dt 2 = — kx 
Since dx/dt — v, this can also be written as 



m li = ~ KX ° r m dxlt = ~ KX > 



dv dv dx 

,-— = — K x or 
dt 

Integration yields fanv 2 + fax 2 = E. 



dv 

i.e. mv-=- = —kx 

dx 



FORCED VIBRATIONS AND RESONANCE 

4.18. Derive the steady-state solution (27) corresponding to the differential equation (24) 
on page 89. 

The differential equation is ••,„•, * 

x + 2ya? + u 2 x = / cos at 



Consider a particular solution having the form 

X = Ci cos at + c 2 sin at 



(1) 
(*) 



100 THE SIMPLE HARMONIC OSCILLATOR AND THE SIMPLE PENDULUM [CHAP. 4 

where c x and c 2 are to be determined. Substituting (2) into (1), we find 

(-a 2 *?! + 2yaC 2 + u 2 c t ) cos at + (—a 2 c 2 — 2yac 1 + w 2 c 2 ) sin at = / cos at 
from which -« 2 Cl + 2yac 2 + w 2 Cl = / , - a 2 c 2 - 2yac x + « 2 c 2 = (3) 

or (a 2 ~ w 2 )cj - 2yac 2 = -/„, 2yaC x + (a 2 - <* 2 )c 2 = (4) 

Solving these simultaneously, we find 

— /o (<o 2 — a 2 ) _ 2/ yw 

Cl ~ (a 2 - <o 2 ) 2 + 4y 2 <o 2 ' C2 ~ (a 2 - <o 2 ) 2 + 4y 2 a 2 (5) 

Thus (2) becomes 

_ /o [(" 2 — « 2 ) cos at + 2ya sin at] 
* ~ (a 2 - co 2 ) 2 + 4y 2 a 2 ( e ) 

Now by Problem 4.2, page 92, 

(<o 2 - a 2 ) cos at + 2ya sin at - V(« 2 ~ a 2 ) 2 + 4y 2 a 2 cos (at - <f>) (7) 

where tan <p = 2ya/(a 2 - <o 2 ), ^ <j> ^ v. Using (7) in (5), we find as required 

x = — cos (at — 0) 

V(« 2 ~ <o 2 ) 2 + 4y 2 a 2 

4.19. Prove (a) that the amplitude in Problem 4.18 is a maximum where the resonant fre- 
quency is determined from a = \/«> 2 - 2y 2 and (b) that the value of this maximum 
amplitude is / /(2y\/ w 2 -y 2 ). 

Method 1. The amplitude in Problem 4.18 is 

/o/vV - <o 2 ) 2 + 4y 2 « 2 (1) 

It is a maximum when the denominator [or the square of the denominator] is a minimum. To 
find this minimum, write 

(a 2 - <o 2 ) 2 + 4y 2 a 2 = a 4 - 2(<o 2 - 2y 2 )a 2 + w 4 

= a 4 - 2(w 2 - 2y 2 )a 2 + (a> 2 - 2y 2 ) 2 + <o 4 - (<o 2 - 2y 2 ) 2 

= [a 2 - (« 2 - 2y 2 )] 2 + 4y 2 (a> 2 - y 2 ) 

This is a minimum where the first term on the last line is zero, i.e. when a 2 = w 2 — 2y 2 , and the 
value is then 4y 2 (w 2 — y 2 ). Thus the value of the maximum amplitude is given from (i) by 

/o/^yV^Y 2 )- 

Method 2. The function U — (a 2 — w 2 ) 2 + 4y 2 a 2 has a minimum or maximum when 

$T- = 2(a 2 - a> 2 )2a + 8y 2 a = or a(a 2 - <o 2 + 2y 2 ) = 
da 

i.e. a = 0, a — Vw 2 — 2y 2 where y 2 < ^w 2 . Now 

dPU/da 2 = 12a 2 - 4<o 2 + 8y 2 



For « = 0, d 2 U/da 2 = -4(<o 2 - 2y 2 ) < 0. For a = Vw 2 - 2y 2 , d 2 U/da 2 = 8(a> 2 - 2y 2 ) > 0. Thus 
a = V" 2 — 2y 2 gives the minimum value. 



4.20. (a) Obtain the solution (33), page 90, for the case where there is no damping and the 
impressed frequency is equal to the natural frequency of the oscillation, (b) Give a 
physical interpretation. 

(a) The case to be considered is obtained by putting y = or (3 = and a = a in equations 
{23) or (24), page 89. We thus must solve the equation 

x + a 2 x = / cos tot (■*) 



CHAP. 41 THE SIMPLE HARMONIC OSCILLATOR AND THE SIMPLE PENDULUM 



101 



To find the general solution of this equation we add the general solution of 

x + u 2 x = (*) 

to a particular solution of (1). 

Now the general solution of (2) is 

x — A cos ut + B sin ut (#) 

To find a particular solution of (1) it would do no good to assume a particular solution of the form 

x = c t cos ut + c 2 sin ut w 

since when we substitute (-4) [which is identical in form to {3)] into the left side of {1), we would 
get zero. We must therefore modify the form of the assumed particular solution (4). As seen 
in Appendix C, the assumed particular solution has the form 

x = t{Ci cos ut + c 2 sin ut) w) 

To see that this yields the required particular solution, let us differentiate (5) to obtain 



x — t(— «c x sin ut + uc 2 cos ut) + (c x cos ut + c 2 sin ut) 

x = t(— u 2 c x cos ut — u 2 c 2 sin ut) + 2(— uc t sin ut + uc 2 cos ut) 

Substituting (5), (6) and (7) into (1), we find after simplifying 

— 2uc t sin ut + 2uc 2 cos ut = f cos ut 

from which c x = and c 2 = f /2u. Thus the re- 
quired particular solution (5) is x — (f /2u)t sin ut. 
The general solution of (1) is therefore 

x = A cos wt + B sin <ot + {fJ2u)t sin tot (5) 



(6) The constants A and 5 in (8) are determined 
from the initial conditions. Unlike the case with 
damping, the terms involving A and B do not become 
small with time. However, the last term involving t 
increases with time to such an extent that the spring 
will finally break. A graph of the last term shown 
in Fig. 4-15 indicates how the oscillations build up 
in magnitude. 



(6) 
(7) 




Fig. 4-15 



4.21. A vertical spring has a stiffness factor equal to 3 lb wt per ft. At £ = a force 
given in lb wt by F(t) = 12 sin4£, t ^ is applied to a 6 lb weight which hangs in 
equilibrium at the end of the spring. Neglecting damping, find the position of the 
weight at any later time t. 

Using the method of Problem 4.7, we have by Newton's second law, 

6 d 2 z 



32 dt 2 
d?z 



= -Sz + 12 sin At 



^-| + 162 = 64 sin 4t 



(1) 



Solving ' z = A cos At + B sin 4t - 8t cos At 

When t = 0, z = and dz/dt = 0; then A = 0, B = 2 and 

z - 2 sin At - 8t cos At (2) 

As t gets larger the term —St cos At increases numerically without bound, and physically the 
spring will ultimately break. The example illustrates the phenomenon of resonance. Note that 
the natural frequency of the spring (4/2*r = 2/w) equals the frequency of the impressed force. 



102 



THE SIMPLE HARMONIC OSCILLATOR AND THE SIMPLE PENDULUM [CHAP. 4 



4.22. Work Problem 4.21 if F(t) = 30 cos Gt, t^O. 

In this case the equation (1) of Problem 4.21 becomes 

d 2 z/dt 2 + 16z = 160 cos 5t (1) 

and the initial conditions are 

z = 0, dz/dt - at t - (2) 

The general solution of (1) is 

z - A cos4£ + B sin 4* — 8 cos 6t (8) 

Using conditions (2) in (5), we find A = 8, £ = and 

2 = 8(cos 4t - cos 6t) = 8{cos(5«-«) - cos(5£ + £)} = 16 sin t sin 5t 

The graph of z vs. t is shown by the heavy curve of Fig. 4-16. The dashed curves are the curves 
z = ±16 sin t obtained by placing sin 5t = ±1. If we consider that 16 sin t is the amplitude of 
sin5«, we see that the amplitude varies sinusoidally. The phenomenon is known as amplitude 
modulation and is of practical importance in communications and electronics. 




Fig. 4-16 



THE SIMPLE PENDULUM 

4.23. Determine the motion of a simple pendulum of length I and mass m assuming small 
vibrations and no resisting forces. 

Let the position of m at any time be determined by s, 
the arclength measured from the equilibrium position 
[see Fig. 4-17]. Let o be the angle made by the pendulum 
string with the vertical. 

If T is a unit tangent vector to the circular path of 
the pendulum bob m, then by Newton's second law 



m dP T = 



-mg sin e T 



(1) 



or, since s = le, 



<&8 _ g . 



For small vibrations we can replace sin e by B so that 
to a high degree of accuracy equation (2) can be replaced by 

dH 



dP 



+ 5< = ° 



(3) 



which has solution 



e = A coay/gTlt + B sinVFT^* 
Taking as initial conditions $ = $ , de/dt = at t = 0, we find A = 

e = e cos VoU t 
From this we see that the period of the pendulum is 2iry/l/g. 




CHAP. 4] THE SIMPLE HARMONIC OSCILLATOR AND THE SIMPLE PENDULUM 103 

4.24. Show how to obtain the equation (2) for the pendulum of Problem 4.23 by using 
the principle of conservation of energy. 

We see from Fig. 4-17 that OA = OC - AC = I - I cos e - 1(1 - cos o). Then by the conserva- 
tion of energy [taking the reference level for the potential energy as a horizontal plane through 
the lowest point O] we have 

Potential energy at B + Kinetic energy at B = Total energy = E = constant 

mgl(l ~ cos e) + ^m(ds/dt) 2 = E (1) 

Since s — Id, this becomes 

mgl{\ - cos e) + \ml 2 (de/dt) 2 = E (2) 

Differentiating both sides of (2) with respect to t, we find 

mgl sin e h + ml 2 = or V + (g/l) sin — 

in agreement with equation (2) of Problem 4.23. 

4.25. Work Problem 4.23 if a damping force proportional to the instantaneous velocity is 
taken into account. 

In this case the equation of motion (1) of Problem 4.23 is replaced by 

d 2 s ds d 2 s B ds 

m d^ T = -mgsmeT - B Tt T or w = -g sin e - - ^ 

Using s = le and replacing sin o hy d for small vibrations, this becomes 



dt* m dt I 



Three cases arise: 



Case 1. p 2 /4m 2 < g/l 

e - e~-^ t/2m (A cosut + B smut) where w = y/g/l - B 2 lkm 2 

This is the case of damped oscillations or underdamped motion. 

Case 2. B 2 /4m 2 = g/l 

e = e~^ t/2m (A+Bt) 

This is the case of critically damped motion. 

Case 3. B 2 /4m 2 > g/l 

e = e -3t/2m(^ e \t + J5 e -xt) where X = V)8 2 /4m 2 - g/l 

This is the case of overdamped motion. 

In each case the constants A and B can be determined from the initial conditions. In Case 1 
there are continually decreasing oscillations. In Cases 2 and 3 the pendulum bob gradually returns 
to the equilibrium position without oscillation. 



THE TWO AND THREE DIMENSIONAL HARMONIC OSCILLATOR 

4.26. Find the potential energy for (a) the two dimensional and (b) the three dimensional 
harmonic oscillator. 

(a) In this case the force is given by 

F = — k^xi — K 2 yj 

Since VXF = 0, the force field is conservative. Thus a potential does exist, i.e. there exists 
a function V such that F = — VV. We thus have 

dV . dV . av, 



104 THE SIMPLE HARMONIC OSCILLATOR AND THE SIMPLE PENDULUM [CHAP. 4 

from which dV/dx = Kl x, dV/dy - K 2 y, dV/dz = or 

V = 1^*2 + ^ K2 y2 
choosing the arbitrary additive constant to be zero. This is the required potential energy. 

(b) In this case we have F = — k x xi — K 2 y] - K S zk which is also conservative since V X F = 0. 
We then find as in part (a), dV/dx - k x x, 8V/dy = K 2 y, dV/dz = k 3 z from which the required 
potential energy is 

V = fax* + 1k 2 2/2 + fa z 2 



4.27. A particle moves in the xy plane in a force field given by F = - K xi - K yj. Prove that 
in general it will move in an elliptical path. 

If the particle has mass ra, its equation of motion is 

m-jp = F = -kxi - K yj (l) 

d oc ct - i/ 

or, since r = xi + yj, m^i + m-~j = - K xi - K yj 

ct ! ic d^ij 

Then m — = - KX , m-^ = - K y (2) 

These equations have solutions given respectively by 

x = A x cos yj kItyi t + A 2 sin V^M t, y — B t cos y/ n/m t + B 2 sin V '/c/m t (3) 

Let us suppose that at t — the particle is located at the point whose position vector is 
r = ai + bj and movin g with ve locity dr/dt = v x i + v 2 }. Using these conditions, we find A t = a, 
B t = b, A 2 — Vis/mU, B 2 = v 2 y/m/K and so 

x — a cos wt + c sin wt, y = b cos ut + d sin at (4) 

where c = v{\fmTic, d = v 2 y/m/K. Solving for sin ut and cosut in (4) we find, if ad ¥= be, 

dx — cy . , ay — bx 

cos ut = —3 — r- , sin ut = — j — j— 

ad — be ad— be 

Squaring and adding, using the fact that cos 2 ut + sin 2 w £ = 1, we find 

(dx - cy)* + (ay - bx) 2 = (ad- be) 2 

or (62 + d 2 )x 2 - 2(cd + ab)xy + (a* + c 2 )y* = (ad - be) 2 (5) 

Now the equation 

Ax 2 + Bxy + Cy 2 = D where A>0, C > 0, D > 

is an ellipse if B 2 -4AC < 0, a parabola if B 2 — 4AC = 0, and a hyperbola if B 2 -4AC>0. 
To determine what (5) is, we see that A = b 2 + d 2 , B = -2(cd + ab), C = a 2 + c 2 so that 

B 2 - 4AC = 4(cd + ab) 2 - 4(b 2 + d?)(a 2 + c 2 ) = -4(ad - be) 2 < 

provided ad ¥= be. Thus in general the path is an ellipse, and if A - C it is a circle. If ad = be 
the ellipse reduces to the straight line ay = bx. 



MISCELLANEOUS PROBLEMS 

4.28. A cylinder having axis vertical floats in a liquid of density a. It is pushed down 
slightly and released. Find the period of the oscillation if the cylinder has weight W 
and cross sectional area A. 



CHAP. 4] THE SIMPLE HARMONIC OSCILLATOR AND THE SIMPLE PENDULUM 



105 



Let RS, the equilibrium position of the cylinder, be distant 
z from the liquid surface PQ at any time t. By Archimedes' 
principle, the buoyant force on the cylinder is (Az)a. Then by 
Newton's second law, 

W d?z 
9 d& 



rrs" = —Azo 



d?z , gAo 
dt* W z 



= 



Solving, 



z = c t cos yJgAolW t + c 2 sin ^JgAaJW t 




and the period of the oscillation is 2vyW/gAo. 



Fig. 4-18 



4.29. Show that if the assumption of small vibrations is not made, then the period of a 
simple pendulum is 



41 S. 



d<f> 



VI ~k 2 sin 2 */, 



where k = sin (<9 /2) 



The equation of motion for a simple pendulum if small vibrations are not assumed is 

[equation (34), page 91] 

d 2 e a . 



Let de/dt = u. Then 



and (1) becomes 



Integrating (2) we obtain 





dt* 








dH 
d& 


_ du 
~ dt 


= 


du de _ 

de dt ~ 


du 

de 




du 
de 


= 


-fsin* 





(1) 



-r- — y cos + c 



Now when e = e , u = so that c = ~(g/l) cos e . Thus (3) can be written 

u 2 = (2g/l)(cos e - cos e ) or de/dt = ±^(2g/l)(cose - cos0 o ) 



(S) 



U) 



If we restrict ourselves to that part of the motion where the bob goes from e = e to = 0, 
which represents a time equal to one fourth of the period, then we must use the minus sign in (4) 

so that it becomes 

de/dt = - y/(2g/l)(cos o — cos e ) 



Separating the variables and integrating, we have 

• ■ -us 



de 



Vcos e — cos e 

Since t = at e = e and * = P/4 at e = 0, where P is the period, 

de 



[T C e ° 



Vcos e — cos e 



(5) 



Making use of the trigonometric identity cos e = 2 sin 2 (*/2) — 1, with a similar one replacing 
e by e , (5) can be written 



P = 2 



vir 



d* 



Now let 



Vsin 2 (0 o /2) - sin 2 (<?/2) 
sin (e/2) = sin (<? /2) sin 



(7) 



106 THE SIMPLE HARMONIC OSCILLATOR AND THE SIMPLE PENDULUM [CHAP. 4 

Then taking the differential of both sides, 

\ cos (0/2) de — sin (e /2) cos <p d<f> 

or calling k — sin (e /2), 

2 sin (e /2) cos <p d<f> 

de — — 

Vl - k 2 sin 2 <f> 

Also from (7) we see that when e = 0, <f> = 0; and when e = e , <j> = jt/2. Hence (6) becomes, 
as required, 

_ ,7r/2 d±_ 

\ Vl - k 2 sin 2 



* - <VJr-=2= 



Note that if we have small vibrations, i.e. if k is equal to zero very nearly, then the period (8) 
becomes 



p = 4 VJC^ = ^VI 



(9) 

as we have already seen. 

The integral in (8) is called an elliptic integral and cannot be evaluated exactly in terms of 
elementary functions. The equation of motion of the pendulum can be solved for in terms 
of elliptic functions which are generalizations of the trigonometric functions. 



4.30. Show that period given in Problem 4.29 can be written as 

The binomial theorem states that if \x\ < 1, then 

<!+*).> = i + p. + &=$* + "Vffr" - + ■•■ 

If p = ■—■!■, this can be written 

d + aO-i/2 = i -\ x + ^f*» - f^frf* 3 + ••• 

Letting x = —k 2 sin 2 <£ and integrating from to v/2, we find 

/*ir/2 djp 

P - AyjUg I r ,o • o^ 

J yl — k 2 sin 2 tf> 

X ir/2 f 1 1 • 3 1 

-j 1 + | A; 2 sin 2 ^ + 2T4 ** sin4 * + ' " ' | d(f> 

= 2.vw{i + (i)'*. + (£!)'*• + (Hfcf)'"' + ••■} 

where we have made use of the integration formula 

r /2 • a, a l*3'5---(2w-l) ff 

I sin 2 "0^ = 2 .4.6---(2n) 2 

The term by term integration is possible since |fc| < 1. 

4.31. A bead of mass m is constrained to move on a frictionless wire in the shape of a 
cycloid [Fig. 4-19 below] whose parametric equations are 

x = a(cj> - sin </>), y = a(l — cos </>) CO 

which lies in a vertical plane. If the bead starts from rest at point O, (a) find the 
speed at the bottom of the path and (b) show that the bead performs oscillations 
with period equivalent to that of a simple pendulum of length 4a. 



CHAP. 4] THE SIMPLE HARMONIC OSCILLATOR AND THE SIMPLE PENDULUM 



107 



(a) 



4.32. 



Let P be the position of the bead at any time t 
and let s be the arclength along the cycloid meas- 
ured from point O. 

By the conservation of energy, measuring 
potential energy relative to line AB through the 
minimum point of the cycloid, we have 

P.E. at P + K.E. at P = P.E. at O + K.E. at 
mg(2a-y) + im(ds/dt) 2 = mg(2a) + (2) 



Thus 



(ds/dt) 2 = 2gy or 




ds/dt = 



At the lowest point y — 2a the speed is v = -\j2g(2a) = 2y/ga. 

(b) From part (a), (ds/dt) 2 = 2gy. But 

{ds/dt) 2 = (dx/dt) 2 + (dy/dt) 2 = a 2 (l - cos <f>) 2 $ 2 + a 2 sin 2 <f> $ 2 = 2a 2 (l - cos <f>)]> 2 

Then 2a 2 (l - cos <f>)^> 2 = 2ga{l — cos </>) or $ 2 = g/a. Thus 

dtp/dt = yfgja, and <f> = y/g/at + c x (4) 

When <£ = 0, t — 0; when .£ = 2w, t = P/2 where P is the period. Hence from the second 

equation of (t), 

P = Airya/g — 2w-y±a/g 

and the period is the same as that of a simple pendulum of length I = 4a. 
For some interesting applications see Problems 4.86-4.88. 



A particle of mass m is placed on the inside 
of a smooth paraboloid of revolution having 
equation cz = x 2 + y 2 at a point P which is at 
height H above the horizontal [assumed as the 
xy plane]. Assuming that the particle starts 
from rest, (a) find the speed with which it 
reaches the vertex 0, (b) find the time r taken, 
and (c) find the period for small vibrations. 

It is convenient to choose the point P in the yz 
plane so that x = and cz = y 2 . By the principle 
of conservation of energy we have if Q is any point 
on the path PQO, 




Fig. 4-20 



P.E. at P + K.E. at P = P.E. at Q + K.E. at Q 
mgH + £m(0) 2 = mgz + $m(ds/dt) 2 

where 8 is the arclength along OPQ measured from O. Thus 

(ds/dt) 2 = 2g(H-z) 

or ds/dt = — s/2g(H — z) 

using the negative sign since s is decreasing with t. 

(a) Putting z = 0, we see that the speed is y]2gH at the vertex. 



(1) 



(6) We have, since x = and cz = y 2 , 

fds\ 2 fdx\* fdyY , fdz\ 2 _ fdyY.W/dy 

\dtj \dtj \dtj "*" \dt) ~ \dtj "*" c 2 \dt 

Thus (I) can be written (1 + 4y 2 /c 2 )(dy/dt) 2 = 2g(H - y 2 /c). Then 

^7 = - y/2gc -= or 

Vc 2 + 4y2 



= <*+Wt) 



dt 



V"c2 + 4i/2 

V2^c dt = , dy 

yfcH - y* 



108 THE SIMPLE HARMONIC OSCILLATOR AND THE SIMPLE PENDULUM [CHAP. 4 

Integrating, using the fact that z = H and thus y = V^H at t = while at t = t, y - 0, 

we have 

f T , f° Vc 2 + 4y2 1 r \^H Vc 2 + % 2 

I —y2gc dt = I ■ dy or r = — F == I / <fy 

•A) J^ ^/ c # ~ 2/ 2 v20C J VcH - 2/2 

Letting y = yJcH cos (9, the integral can be written 

T = -^— I Vc 2 + 4c# cos 2 9 de = — I Vc 2 + 4cH - AcH sin 2 d* 

and this can be written , v/2 

r = j£±i«J y/T^W^Jde (3) 

where fc = y/AH/{c + AH) < 1 (4) 

The integral in (5) is an elliptic integral and cannot be evaluated in terms of elementary 
functions. It can, however, be evaluated in terms of series [see Problem 4.119]. 

(c) The particle oscillates back and forth on the inside of the paraboloid with period given by 

p = 4t = 4 J c + m r /2 VI - fe 2 sin 2 » cZg (5) 

For small vibrations the value of & given by &) can be assumed so small so as to be zero 
for practical purposes. Hence (5) becomes 

P = 2w^fJc + A~H)j29 

The length of the equivalent simple pendulum is I - ^(c + AH). 



Supplementary Problems 

SIMPLE HARMONIC MOTION AND THE SIMPLE HARMONIC OSCILLATOR 

4.33. A particle of mass 12 gm moves along the x axis attracted toward the point O on it by a force 
in dynes which is numerically equal to 60 times its instantaneous distance x cm from O. If the 
particle starts from rest at x = 10, find the (a) amplitude, (6) period and (c) frequency of 
the motion. Ans. (a) 10 cm, (6) 2jt/Vb sec, (c) Vb/2jt vib/sec 

4.34. (a) If the particle of Problem 4.33 starts at x = 10 with a speed toward O of 20 cm/sec, determine 
its amplitude, period and frequency. (6) Determine when the particle reaches O for the first time. 
Ans. (a) Amplitude = 6^ cm, period = 2^/V5 sec, frequency = V5/2*- vib/sec; (b) 0.33 sec 

4 35 A particle moves on the x axis attracted toward the origin O on it with a force proportional 
to its instantaneous distance from O. If it starts from rest at x = 5 cm and reaches ; x - 2.5 cm 
for the first time after 2 sec, find (a) the position at any time t after it starts, (6) the speed 
at x = 0, (c) the amplitude, period and frequency of the vibration, (d) the maximum acceleration, 
(e) the maximum speed. 

Ans. (a) a = 5cosM/6); (6) 5W6 cm/sec; (c) 5 cm, 12 sec, 1/12 vib/sec; (d) 5^/36 cm/sec 2 ; 
(e) 5?r/6 cm/sec 

4 36 If a particle moves with simple harmonic motion along the x axis, prove that (a) the acceleration 
is numerically greatest at the ends of the path, (6) the velocity is numerically greatest in the 
middle of the path, (c) the acceleration is zero in the middle of the path, (d) the velocity is zero 
at the ends of the path. 

4 37 A particle moves with simple harmonic motion in a straight line. Its maximum speed is 20 ft/sec 
' ' and its maximum acceleration is 80 ft/sec 2 . Find the period and frequency of the motion. 
Ans. v/2 sec, 2/v vib/sec 



CHAP. 4] THE SIMPLE HARMONIC OSCILLATOR AND THE SIMPLE PENDULUM 109 

4.38. A particle moves with simple harmonic motion. If its acceleration at distance D from the 
equilibrium position is A, prove that the period of the motion is 2tt^D/A. 

4.39. A particle moving with simple harmonic motion has speeds of 3 cm/sec and 4 cm/sec at distances 
8 cm and 6 cm respectively from the equilibrium position. Find the period of the motion. 
Ans. Av sec 

4.40. An 8 kg weight placed on a vertical spring stretches it 20 cm. The weight is then pulled down 
a distance of 40 cm and released, (a) Find the amplitude, period and frequency of the oscillations. 
(b) What is the position and speed at any time? 

Ans. (a) 40 cm, 2ir/l sec, 7/2tt vib/sec 

(b) x = 40 cos It cm, v = —280 sin It cm/sec 

4.41. A mass of 200 gm placed at the lower end of a vertical spring stretches it 20 cm. When it is in 
equilibrium, the mass is hit and due to this goes up a distance of 8 cm before coming down again. 
Find (a) the magnitude of the velocity imparted to the mass when it is hit and (6) the period of 
the motion. Ans. (a) 56 cm/sec, (6) 2jt/7 sec 

4.42. A 5 kg mass at the end of a spring moves with simple harmonic motion along a horizontal straight 
line with period 3 sec and amplitude 2 meters, (a) Determine the spring constant. (6) What is the 
maximum force exerted on the spring? 

Ans. (a) 1140 dynes/cm or 1.14 newtons/meter 

(6) 2.28 X 10 5 dynes or 2.28 newtons 

4.43. When a mass M hanging from the lower end of a vertical spring i s set into motion, it oscillates 
with period P. Prove that the period when mass m is added is Py/l + m/M. 

THE DAMPED HARMONIC OSCILLATOR 

4.44. (a) Solve the equation d 2 x/dt 2 + 2 dx/dt + 5x - subject to the conditions x = 5, dx/dt = -3 
at t — and (6) give a physical interpretation of the results. 

Ans. (a) x — J. e -t(10 cos 2* -5 sin 2i) 

4.45. Verify that the damping force given by equation (2) of Problem 4.11 is correct regardless of the 
position and velocity of the particle. 

4.46. A 60 lb weight hung on a vertical spring stretches it 2 ft. The weight is then pulled down 3 ft 
and released, (a) Find the position of the weight at any time if a damping force numerically 
equal to 15 times the instantaneous speed is acting. (6) Is the motion oscillatory damped, over- 
damped or critically damped? Ans. (a) x - 3e-«(4£+l), (6) critically damped 

4.47. Work Problem 4.46 if the damping force is numerically 18.75 times the instantaneous speed. 
Ans. (a) x = 4e~ 2t — e~ 8t , (b) overdamped 

4.48. In Problem 4.46, suppose that the damping force is numerically 7.5 times the instantaneous speed. 
(a) Prove that the motion is damped oscillatory. (6) Find the amplitude, period and frequency of 
the oscillations, (c) Find the logarithmic decrement. 

Ans. (b) Amplitude = 2^3 e~^ ft, period = ir/yfz sec, frequency = yfzh vib/sec; (c) 2^/V3 

4.49. Prove that the logarithmic decrement is the time required for the maximum amplitude during 
an oscillation to reduce to 1/e of this value. 

4.50. The natural frequency of a mass vibrating on a spring is 20 vib/sec, while its frequency with 
damping is 16 vib/sec. Find the logarithmic decrement. Ans. 3/4 

4.51. Prove that the difference in times corresponding to the successive maximum displacements of a 
dampe d harmon ic oscillator with equation given by (12) of page 88 is constant and equal to 
4n-m/\/4/cm — /3 2 . 

4.52. Is the difference in times between successive minimum displacements of a damped harmonic 
oscillator the same as in Problem 4.51? Justify your answer. 



110 THE SIMPLE HARMONIC OSCILLATOR AND THE SIMPLE PENDULUM [CHAP. 4 

FORCED VIBRATIONS AND RESONANCE 

4.53. The position of a particle moving along the x axis is determined by the equation d 2 x/dt 2 + Adxldt + 
8* = 20 cos 2t. If the particle starts from rest at x = 0, find (a) x as a function of t, (b) the 
amplitude, period and frequency of the oscillation after a long time has elapsed. 

Ans. (a) x = cos 2t + 2 sin 2t - e~ 2t (cos 2t + 3 sin 2t) 

(b) Amplitude = yfh, period = v, frequency = 1/v 

4.54. (a) Give a physical interpretation to Problem 4.53 involving a mass at the end of a vertical spring. 
(b) What is the natural frequency of such a vibrating spring? (c) What is the frequency of the 
impressed force? Ans. (b) y/2/ir, (c) 1/tt 

4.55. The weight on a vertical spring undergoes forced vibrations according to the equation 
d 2 x/dt 2 + 4x = 8 sin at where x is the displacement from the equilibrium position and w > is a 
constant. If at t = 0, x = and dx/dt = 0, find (a) x as a function of t, (b) the period of 
the external force for which resonance occurs. 

Ans. (a) x = (8 sin ut — 4w sin 2t)/(4 — <o 2 ) if w ¥^ 2; x = sin 2t — 2t cos 2t if w = 2 
(6) w = 2 or period = it 

4.56. A vertical spring having constant 17 lb wt per ft has a 32 1b weight suspended from it. Ah 
external force given as a function of time t by F(t) = 65 sin4t, i§0 is applied. A damping 
force given numerically in lb wt by 2v, where v is the instantaneous speed of the weight in ft/sec, 
is assumed to act. Initially the weight is at rest at the equilibrium position, (a) Determine the 
position of the weight at any time. (6) Indicate the transient and steady-state solutions, giving 
physical interpretations of each, (c) Find the amplitude, period and frequency of the steady-state 
solution. [Use g = 32 ft/sec 2 .] 

Ans. (a) x = 4e -t cos4t + sin 4t — 4 cos 4* 

(6) Transient, 4e -t cos4t; steady-state, sin4i — 4 cos4t 

(c) Amplitude = y/Vf ft, period = v/2 sec, frequency = 2/tr vib/sec 

4.57. A spring is stretched 5 cm by a force of 50 dynes. A mass of 10 gm is placed on the lower end 
of the spring. After equilibrium has been reached, the upper end of the spring is moved 
up and down so that the external force acting on the mass is given by F(t) = 20 cos at, t§0. 
(a) Find the position of the mass at any time, measured from its equilibrium position. (6) Find the 
value of (o for which resonance occurs. 

Ans. (a) x = (20 cos ut)/(l - u 2 ) - 20 cos t, (b) u = 1 

4.58. A periodic external force acts on a 6 kg mass suspended from the lower end of a vertical spring 
having constant 150 newtons/meter. The damping force is proportional to the instantaneous speed 
of the mass and is 80 newtons when the speed is 2 meters/sec. Find the frequency at which 
resonance occurs. Ans. 5/6v vib/sec 



THE SIMPLE PENDULUM 

4.59. Find the length of a simple pendulum whose period is 1 second. Such a pendulum which registers 
seconds is called a seconds pendulum. Ans. 99.3 cm or 3.26 ft 

4.60. Will a pendulum which registers seconds at one location lose or gain time when it is moved to 
another location where the acceleration due to gravity is greater? Explain. 

Ans. Gain time 

4.61. A simple pendulum whose length is 2 meters has its bob drawn to one side until the string makes 
an angle of 30° with the vertical. The bob is then released, (a) What is the speed of the bob as 
it passes through its lowest point? (6) What is the angular speed at the lowest point? (c) What 
is the maximum acceleration and where does it occur? 

Ans. (a) 2.93 m/sec, (6) 1.46 rad/sec, (c) 2 m/sec 2 



CHAP. 4] THE SIMPLE HARMONIC OSCILLATOR AND THE SIMPLE PENDULUM HI 

4.62. Prove that the tension in the string of a vertical simple pendulum of length I and mass m is given 
by mg cos e where a is the instantaneous angle made by the string with the vertical. 

4.63. A seconds pendulum which gives correct time at a certain location is taken to another location 
where it is found to lose T seconds per day. Determine the gravitational acceleration at the second 
location. Ans. g(l — T/86,400) 2 where g is the gravitational acceleration at the first location 

4.64. What is the length of a seconds pendulum on the surface of the moon where the acceleration due 
to gravity is approximately 1/6 that on the earth? Ans. 16.5 cm 

4.65. A simple pendulum of length I and mass m hangs vertically from a fixed point O. The bob is given 
an initial horizontal velocity of magnitude v . Prove that the arc through which the bob swings 
in one period has a length given by 41 cos -1 (1 — v 2 /2gl) 

4.66. Find the minimum value of v in Problem 4.65 in order that the bob will make a complete 
vertical circle with center at 0. Ans. 2~\fgl 

THE TWO AND THREE DIMENSIONAL HARMONIC OSCILLATOR 

4.67. A particle of mass 2 moves in the xy plane attracted to the origin with a force given by 
F = — 18*i — 50?/j. At t = the particle is placed at the point (3, 4) and given a velocity 
of magnitude 10 in a direction perpendicular to the x axis, (a) Find the position and velocity of 
the particle at any time. (6) What curve does the particle describe? 

Ans. (a) r = 3 cos 3t i + [4 cos ht + 2 sin 5«] j, v = — 9 sin St i + [10 cos 5t - 20 sin 5t] j 

4.68. Find the total energy of the particle of Problem 4.67. Ans. 581 

4.69. A two dimensional harmonic oscillator of mass 2 has potential energy given by V = S(x 2 + 4y 2 ). 
If the position vector and velocity of the oscillator at time t = are given respectively by 
r = 2i — j and v = 4i + 8j, (a) find its position and velocity at any time t > and (6) deter- 
mine the period of the motion. 

Ans. (a) r = (2 cos At + sin 4t)i + (sin St - cos 8«)j, v = (4 cos 4t - 8 sin 4*)i + (8 cos 8i + 8 sin 8*)j 
(6) s-/8 

4.70. Work Problem 4.69 if V = 8(x 2 + 2y 2 ). Is there a period defined for the motion in this case? 
Explain. 

4.71. A particle of mass m moves in a 3 dimensional force field whose potential is given by 
V = ^k(x 2 + 4y 2 + 16z 2 ). (a) Prove that if the particle is placed at an arbitrary point in space 
other than the origin, then it will return to the point after some period of time. Determine this 
time. (6) Is the velocity on returning to the starting point the same as the initial velocity? Explain. 

4.72. Suppose that in Problem 4.71 the potential is V = %k(x 2 + 2y 2 + 5« 2 ). Will the particle return 
to the starting point? Explain. 

MISCELLANEOUS PROBLEMS 

4.73. A vertical spring of constant k having natural length I is supported at a fixed point A. A mass m 
is placed at the lower end of the spring, lifted to a height h below A and dropped. Prove that 
the lowest point reached will be at a distance below A given by I + mg/ K + y/m 2 g 2 / K 2 + 2mgh/n. 

4.74. Work Problem 4.73 if damping proportional to the instantaneous velocity is taken into account. 

4.75 Given the equation mx + /3x + kx = for damped oscillations of a harmonic oscillator. Prove that 
if E = \mx 2 + \kx 2 , then E = -p'x 2 . Thus show that if there is damping the total energy E 
decreases with time. What happens to the energy lost? Explain. 



112 THE SIMPLE HARMONIC OSCILLATOR AND THE SIMPLE PENDULUM [CHAP. 4 

4.76. (a) Prove that A x cos (at - 4> x ) + A 2 cos (at - <p 2 ) = A cos (at - </>) 

/ A-l sin^t + A 2 sin^> 2 \ 

where A = y/A\ + A\ + 2A 1 A 2 cos (^ - <p 2 ), <fi = ^'(^cos^ + A 2 cos <t>J ' 

(6) Use (a) to demonstrate that the sum of two simple harmonic motions of the same frequency 
and in the same straight line is simple harmonic of the same frequency. 

4.77. Give a vector interpretation to the results of Problem 4.76. 

4.78. Discuss Problem 4.76 in case the frequencies of the two simple harmonic motions are not equal. 
Is the resultant motion simple harmonic? Justify your answer. 

4.79. A particle oscillates in a plane so that its distances x and y from two mutually perpendicular 
axes are given as functions of time t by 

x = A cos (at + X ), y = B cos (at + </> 2 ) 
(a) Prove that the particle moves in an ellipse inscribed in the rectangle defined by x = ±A, 
y — ±b (&) p r ove that the period of the particle in the elliptical path is 2ir/a. 

4.80. Suppose that the particle of Problem 4.79 moves so that 

X = A COS (at + 0i), y — B COS (at + et + 2 ) 

where « is assumed to be a positive constant which is assumed to be much smallertiian a. Vrvve 
that the particle oscillates in slowly rotating ellipses inscribed in the rectangle x - ±A, y - ±B. 

4.81. Illustrate Problem 4.80 by graphing the motion of a particle which moves in the path 

x = 3 cos(2t + n-/4), y = 4 cos (2.4t) 



-nmr- 



m 



K 2 



-^TOKT^- 




4.82. In Fig. 4-21 a mass m which is on a frictionless 
table is connected to fixed points A and B by 
two springs of equal natural length, of negli- 
gible mass and spring constants /q and /c 2 re- ^ 

spectively. The mass m is displaced horizontally A B 

and then released. Prove that the perio d of 

oscillation is given by P = 2v-\/'m/(K 1 + k 2 ). Fig. 4-21 

4.83. A spring having constant k and negligible mass has 
one end fixed at point A on an inclined plane of 
angle a and a mass m at the other end, as indicated 
in Fig. 4-22. If the mass m is pulled down a distance 
x below the equilibrium position and released, find 
the displacement from the equilibrium position at any 
time if (a) the incline is frictionless, (6) the incline 
has coefficient of friction /i. Fig. 4-2 

4.84. A particle moves with simple harmonic motion along the x axis. At times t , 2t Q an dJ*o i* ™ 
located at * ■ = a, b and c respectively. Prove that the period of oscillation is cos -i (a + c )/26* 

4 85. A seconds pendulum giving the correct time at one location is taken to another locati or i where 
it loses 5 minutes per day. By how much must the pendulum rod be lengthened or shortened in 
order to give the correct time? 

4.86. A vertical pendulum having a bob of mass m is sus- 
pended from the fixed point O. As it oscillates, the 
string winds up on the constraint curves OB A [or OC\ 
as indicated in Fig. 4-23. Prove that if curve ABC is a 
cycloid, then the period of oscillation will be the same 
regardless of the amplitude of the oscillations. The pen- 
dulum in this case is called a cycloidal pendulum. The 
curves ODA and OC are constructed to be evolutes of 
the cycloid. [Hint. Use Problem 4.31.] Iig.4-A» 




CHAP. 4] THE SIMPLE HARMONIC OSCILLATOR AND THE SIMPLE PENDULUM 113 

4.87. A bead slides down a frictionless wire located in a vertical plane. It is desired to find the 
shape of the wire so that regardless of where the bead is placed on the wire it will slide under 
the influence of gravity to the bottom of the wire in the same time. This is often called the 
tautochrone problem. Prove that the wire must have the shape of a cycloid. 

[Hint. Use Problem 4.31.] 

4.88. Prove that the curves ODA and OC of Problem 4.86 are cycloids having the same shape as the 
cycloid ABC. 

4.89. A simple pendulum of length I has its point of support moving back and forth on a horizontal line 
so that its distance from a fixed point on the line is A sin ut, t ^ 0. Find the position of the 
pendulum bob at any time t assuming that it is at rest at the equilibrium position at * = 0. 

4.90. Work Problem 4.89 if the point of support moves vertically instead of horizontally and if at 
t — the rod of the pendulum makes an angle O with the vertical. 

4.91. A particle of mass m moves in a plane under the influence of forces of attraction toward fixed 
points which are directly proportional to its instantaneous distance from these points. Prove 
that in general the particle will describe an ellipse. 

4.92. A vertical elastic spring of negligible weight and having its upper end fixed, carries a weight 
W at its lower end. The weight is lifted so that the tension in the spring is zero, and then it is 
released. Prove that the tension in the spring will not exceed 2W. 

4.93. A vertical spring having constant /c has a pan on top of it with 

a weight W on it [see Fig. 4-24]. Determine the largest fre- W 

quency with which the spring can vibrate so that the weight 

will remain in the pan. , 



4.94. A spring has a natural length of 50 cm and a force of 100 dynes 
is required to stretch it 25 cm. Find the work done in stretching 
the spring from 75 cm to 100 cm, assuming that the elastic limit 
is not exceeded so that the spring characteristics do not change. 
Ans. 3750 ergs 

4.95. A particle moves in the xy plane so that its position is given by 
x = Acosw«, y = B cos 2ut. Prove that it describes an arc of a 
parabola. Fig. 4.24 

4.96. A particle moves in the xy plane so that its position is given by x = A cos (u^ + 9^), 
y = Bcos(w 2 t + <f> 2 )- Prove that the particle describes a closed curve or not, according as a t /u 2 is 
rational or not. In which cases is the motion periodic? 

4.97. The position of a particle moving in the xy plane is described by the equations d 2 x/dt 2 = —Ay, 
d 2 y/dt 2 = -4x. At time t = the particle is at rest at the point (6, 3). Find (a) its position 
and (b) its velocity at any later time t. 

4.98. Find the period of a simple pendulum of length 1 meter if the maximum angle which the rod 
makes with the vertical is (a) 30°, (6) 60°, (c) 90°. 

4.99. A simple pendulum of length 3 ft is suspended vertically from a fixed point. At t = the bob is 
given a horizontal velocity of 8 ft/sec. Find (a) the maximum angle which the pendulum rod 
makes with the vertical, (b) the period of the oscillations. 

Ans. (a) cos" 1 2/3 = 41° 48', (6) 1.92 sec 

4.100. Prove that the time averages over a period of the potential energy and kinetic energy of a 
simple harmonic oscillator are equal to 2tt 2 A 2 IP 2 where A is the amplitude and P is the period 
of the motion. 

4.101. A cylinder of radius 10 ft with its axis vertical oscillates vertically in water of density 62.5 lb/ft 3 
with a period of 5 seconds. How much does it weigh? Ans. 3.98Xl0 5 lbwt 

4.102. A particle moves in the xy plane in a force field whose potential is given by V = x 2 + xy + y 2 . 
If the particle is initially at the point (3, 4) and is given a velocity of magnitude 10 in a direction 
parallel to the positive x axis, (a) find the position at any time and (6) determine the period of 
the motion if one exists. 




114 THE SIMPLE HARMONIC OSCILLATOR AND THE SIMPLE PENDULUM [CHAP. 4 

4.103. In Problem 4.96 suppose that a x /a 2 is irrational and that at t — the particle is at the 
particular point (x ,y ) inside the rectangle denned by x = ±A, y — ±B. Prove that the point 
(x ,y ) will never be reached again but that in the course of its motion the particle will come 
arbitrarily close to the point. 

4.104. A particle oscillates on a vertical frictionless cycloid with its vertex downward. Prove that the 
projection of the particle on a vertical axis oscillates with simple harmonic motion. 

4.105. A mass of 5 kg at the lower end of a vertical spring which has an elastic constant equal to 
20 newtons/meter oscillates with a period of 10 seconds. Find (a) the damping constant, (6) the 
natural period and (c) the logarithmic decrement. Ans. (a) 19 nt sec/m, (6) 3.14 sec 

4.106. A mass of 100 gm is supported in equilibrium by two identical 
springs of negligible mass having elastic constant equal to 
50 dynes/cm. In the equilibrium position shown in Fig. 4-25 
the springs make an angle of 30° with the horizontal and are 
100 cm in length. If the mass is pulled down a distance of 
2 cm and released, find the period of the resulting oscillation. 

4.107. A thin hollow circular cylinder of inner radius 10 cm is fixed 
so that its axis is horizontal. A particle is placed on the inner 
frictionless surface of the cylinder so that its vertical distance 
above the lowest point of the inner surface is 2 cm. Find 

(a) the time for the particle to reach the lowest point and 

(b) the period of the oscillations which take place. 

4.108. A cubical box of side a and weight W vibrates vertically in water of density o. Prove that the 
period of vibration is {2ir/a)^<jg/W. 

4.109. A spring vibrates so that its equation of motion is 

md 2 x/dt* + kx = F(t) 
If x = 0, dx/dt - at t = 0, find it as a function of time t. 

i r* i — 

Ans. x = — — I F(u) sin V «/m (t — u) du 



■yrriK ^o 

4.110. Work Problem 4.109 if damping proportional to dx/dt is taken into account. 

4.111. A spring vibrates so that its equation of motion is 

m d 2 x/dt 2 + kx — 5 cos cot + 2 cos 3wt 
If x = 0, x = v at t = 0, (a) find x at any time * and (6) determine for what values of a 
resonance will occur. 

4 112 A vertical spring having elastic constant k carries a mass m at its lower end. At t = the 
spring is in equilibrium and its upper end is suddenly made to move vertically so that its distance 
from the original point of support is given by A sin a t, t § 0. Find (a) the position of the mass w 
at any time and (b) the values of <o for which resonance occurs. 

4.113. (a) Solve d*x/dt* + x = t sin t + cos t where x = 0, dx/dt = at t = 0, and (6) give a physical 
interpretation. 

4.114. Discuss the motion of a simple pendulum for the case where damping and external forces are 
present. 



CHAP. 4] THE SIMPLE HARMONIC OSCILLATOR AND THE SIMPLE PENDULUM 115 

4.115. Find the period of small vertical oscillations of a cylinder of radius a and height h floating 
with its axis horizontal in water of density a. 

4.116. A vertical spring having elastic constant 2 newtons per meter has a 50 gm weight suspended 
from it. A force in newtons which is given as a function of time t by F(t) = 6 cos 4 t, t ^ 
is applied. Assuming that the weight, initially at the equilibrium position, is given an upward 
velocity of 4 m/sec and that damping is negligible, determine the (a) position and (6) velocity 
of the weight at any time. 

4.117. In Problem 4.55, can the answer for « = 2 be deduced from the answer for « ^ 2 by taking 
the limit as u -»• 2? Justify your answer. 

4.118. An oscillator has a restoring force acting on it whose magnitude is — kx — ex 2 where e is small 
compared with k. Prove that the displacement of the oscillator [in this case often called an 
anharmonic oscillator] from the equilibrium position is given approximately by 

A 2 

x = A cos (ut-<f>) + ^—-{cos2(ut — d>)-3} 

OK 

where A and </> are determined from the initial conditions. 

4.119. Prove that if the oscillations in Problem 4.32 are not necessarily small, then the period is given by 



Chapter 5 



CENTRAL FORCES 
and PLANETARY MOTION 






CENTRAL FORCES 

Suppose that a force acting on a particle of 
mass m is such that [see Fig. 5-1]: 

(a) it is always directed from m toward or 
away from a fixed point O, 

(b) its magnitude depends only on the distance 
r from O. 

Then we call the force a central force or central 
force field with as the center of force. In sym- 
bols F is a central force if and only if 

F = f(r)n = f(r)r/r (1) Fig. 5-1 

where ri = r/r is a unit vector in the direction of r. 

The central force is one of attraction toward O or repulsion from O according as 
f(r) < or f(r) > respectively. 

SOME IMPORTANT PROPERTIES OF CENTRAL FORCE FIELDS 

If a particle moves in a central force field, then the following properties are valid. 

1. The path or orbit of the particle must be a plane curve, i.e. the particle moves in 
a plane. This plane is often taken to be the xy plane. See Problem 5.1. 

2. The angular momentum of the particle is conserved, i.e. is constant. See Problem 5.2. 

3. The particle moves in such a way that the position vector or radius vector drawn 
from to the particle sweeps out equal areas in equal times. In other words, the 
time rate of change in area is constant. This is sometimes called the law of areas. 
See Problem 5.6. 



EQUATIONS OF MOTION FOR A PARTICLE 
IN A CENTRAL FIELD 

By Property 1, the motion of a particle in a cen- 
tral force field takes place in a plane. Choosing this 
plane as the xy plane and the coordinates of the par- 
ticle as polar coordinates (r, 6), the equations of mo- 
tion are found to be [see Problem 5.3] 

m(r-r¥) = f{r) (2) 

m(r'e+ 2r9) = (8) 

where dots denote differentiations with respect to 

time t. 




(r,») 



Fig. 5-2 



116 



CHAP. 5] CENTRAL FORCES AND PLANETARY MOTION 117 

From equation (3) we find 

r 2 6 = constant = h (4) 

This is related to Properties 2 and 3 above. 

IMPORTANT EQUATIONS DEDUCED FROM 
THE EQUATIONS OF MOTION 

The following equations deduced from the fundamental equations (2) and (3) often 
prove to be useful. 

1 •• h 2 f(r) 

-!■• T — — = ■ (K\ 

r 3 m * ' 

d 2 u 1 

where u = 1/r. 



d0 2 r\de) r mh 2 ^ 

POTENTIAL ENERGY OF A PARTICLE IN A CENTRAL FIELD 

A central force field is a conservative field, hence it can be derived from a potential. 
This potential which depends only on r is, apart from an arbitrary additive constant, 
given by 

V(r) = -jf(r)dr (g) 

This is also the potential energy of a particle in the central force field. The arbitrary 
additive constant can be obtained by assuming, for example, 7 = at r = or V -* 
as r-* oo. 

CONSERVATION OF ENERGY 

By using (8) and the fact that in polar coordinates the kinetic energy of a particle is 
im^ + r^e 2 ), the equation for conservation of energy can be written 

im(r 2 + r 2 6 2 ) + V(r) = E (P) 

or \m(r 2 + r 2 6 2 ) - f f(r) dr = E (io) 

where E is the total energy and is constant. Using (4), equation (10) can also be written as 

mh 2 Vfdr\ 2 „1 f „ , , 

and also as f ( f2 + S) ~ J ^ dr = E W 

In terms of u = 1/r, we can also write equation (9) as 

f^V ^ . 2 _ 2(E~V) 



KdeJ + u = mh 2 (I 3 ) 



DETERMINATION OF THE ORBIT FROM THE CENTRAL FORCE 

;d, i.e. if f(r) is given, it is possi 
rbit can be obtained in the form 

r = r{0) ( U ) 



If the central force field is prescribed, i.e. if f(r) is given, it is possible to determine 
the orbit or path of the particle. This orbit can be obtained in the form 



118 



CENTRAL FORCES AND PLANETARY MOTION 



[CHAP. 5 



i.e. r as a function of 9, or in the form 

r = r{t), 9 = 9{t) (IS) 

which are parametric equations in terms of the time parameter t. 

To determine the orbit in the form (U) it is convenient to employ equations (6), (7) 
or (11). To obtain equations in the form (15), it is sometimes convenient to use (12) together 
with (A) or to use equations (4) and (5). 

DETERMINATION OF THE CENTRAL FORCE FROM THE ORBIT 

Conversely if we know the orbit or path of the particle, then we can find the correspond- 
ing central force. If the orbit is given by r = r(9) or u = u(9) where u = 1/r, the central 
force can be found from 

mh 2 [d 2 r 2 /dr\ 2 _ 

,de) 



f(r) = 



or 



f(Vu) = 



de* 



-mh 2 u 2 



(16) 



d 2 u 

de 2 



+ u 



(17) 



which are obtained from equations (6) and (7) on page 117. The law of force can also be 
obtained from other equations, as for example equations (9)-(13). 

It is important to note that given an orbit there may be infinitely many force fields for 
which the orbit is possible. However, if a central force field exists it is unique, i.e. it is 
the only one. 



CONIC SECTIONS, ELLIPSE, PARABOLA AND HYPERBOLA 

Consider a fixed point and a fixed line AB distant D from 0, as shown in Fig. 5-3. 
Suppose that a point P in the plane of and AB moves so that the ratio of its distance 
from point to its distance from line AB is always equal to the positive constant € . 

Then the curve described by P is given in 
polar coordinates (r, 9) by 

V 



r — 



(18) 



1 + e COS 9 

See Problem 5.16. 

The point is called a focus, the line AB is 
called a directrix and the ratio e is called the 
eccentricity. The curve is often called a conic 
section since it can be obtained by intersecting 
a plane and a cone at different angles. Three 
possible types of curves exist, depending on the 
value of the eccentricity. 



1 


1 


A 




















^ 






^v 






X 
V 




^- Dir 


p 


\p d 


M 




r/\ 


Focus \^ 


A' \ 







/ 
/ 
/ 
/ 


E 


/ 






/ 


m u » 










S- 
^ 




B 



Fig. 5-3 



Ellipse: « < 1 [See Fig. 5-4 below.] 

If C is the center of the ellipse and CV = CU = a is the length of the semi-major 
axis, then the equation of the ellipse can be written as 



a(l-e 2 ) 

r i + € cos e 



(19) 



Note that the major axis is the line joining the vertices V and U of the ellipse and has 
length 2a. 



CHAP. 5] 



CENTRAL FORCES AND PLANETARY MOTION 



119 



If b is the length of the semi-minor axis 
[CW or CS in Fig. 5-4] and c is the distance 
CO from center to focus, then we have the 
important result 

c = \Za 2 - b 2 = a c (20) 

A circle can be considered as a special case 
of an ellipse with eccentricity equal to zero. 

2. Parabola: £ = 1 [See Fig. 5-5.] 
The equation of the parabola is 



r = 



P 



1 + cos 



(21) 



We can consider a parabola to be a 
limiting case of the ellipse (19) where « -* 1, 
which means that a -» °o [i.e. the major 
axis becomes infinite] in such a way that 
a(l - £ 2 ) = p. 

Hyperbola: e > 1 [See Fig. 5-6.] 

The hyperbola consists of two branches 
as indicated in Fig. 5-6. The branch on the 
left is the important one for our purposes. 
The hyperbola is asymptotic to the dashed 
lines of Fig. 5-6 which are called its asymp- 
totes. The intersection C of the asymptotes 
is called the center. The distance CV = a 
from the center C to vertex V is called the 
semi-major axis [the major axis being the 
distance between vertices V and U by anal- 
ogy with the ellipse]. The equation of the 
hyperbola can be written as 



r = a(< 2 - l) 



1 + e cos 9 



(22) 




Fig. 5-5 




Fig. 5-6 



Various other alternative definitions for conic sections may be given. For example, an 
ellipse can be defined as the locus or path of all points the sum of whose distances from two 
fixed points is a constant. Similarly, a hyperbola can be defined as the locus of all points 
the difference of whose distances from two fixed points is a constant. In both these cases 
the two fixed points are the foci and the constant is equal in magnitude to the length of 
the major axis. 



SOME DEFINITIONS IN ASTRONOMY 

A solar system is composed of a star [such as our sun] and objects called planets which 
revolve around it. The star is an object which emits its own light, while the planets do 
not emit light but can reflect it. In addition there may be objects revolving about the 
planets. These are called satellites. 

In our solar system, for example, the moon is a satellite of the earth which in turn is a 
planet revolving about our sun. In addition there are artificial or man-made satellites 
which can revolve about the planets or their moons. 




120 CENTRAL FORCES AND PLANETARY MOTION [CHAP. 5 

The path of a planet or satellite is called its orbit. The largest and smallest distances 
of a planet from the sun about which it revolves are called the aphelion and perihelion 
respectively. The largest and smallest distances of a satellite around a planet about which 
it revolves are called the apogee and perigee respectively. 

The time for one complete revolution of a body in an orbit is called its period. This is 
sometimes called a sidereal period to distinguish it from other periods such as the period 
of earth's motion about its axis, etc. 

KEPLER'S LAWS OF PLANETARY MOTION 

Before Newton had enunciated his famous laws /^ ^N^pianet 

of motion, Kepler, using voluminous data accumu- 
lated by Tycho Brahe formulated his three laws 
concerning the motion of planets around the sun 
[see Fig. 5-7]. 

1. Every planet moves in an orbit which is an 
ellipse with the sun at one focus. Fig. 5-7 

2. The radius vector drawn from the sun to any planet sweeps out equal areas in 
equal times (the law of areas, as on page 116). 

3. The squares of the periods of revolution of the planets are proportional to the cubes 
of the semi-major axes of their orbits. 

NEWTON'S UNIVERSAL LAW OF GRAVITATION 

By using Kepler's first law and equations (16) or (17), Newton was able to deduce his 
famous law of gravitation between the sun and planets, which he postulated as valid for any 
objects in the universe [see Problem 5.21]. 

Newton's Law of Gravitation. Any two particles of mass mi and m 2 respectively and 
distance r apart are attracted toward each other with a force 

F = _ Gmim 2 ri ^3) 

where G is a universal constant called the gravitational constant. 

By using Newton's law of gravitation we can, conversely, deduce Kepler's laws [see 
Problems 5.13 and 5.23]. The value of G is shown in the table on page 342. 

ATTRACTION OF SPHERES AND OTHER OBJECTS 

By using Newton's law of gravitation, the forces of attraction between large objects 
such as spheres can be determined. To do this, we use the fact that each large object is 
composed of particles. We then apply the law of gravitation to find the forces between 
particles and sum over these forces, usually by methods of integration, to find the resultant 
force of attraction. An important application of this is given in the following 

Theorem 5.1. Two solid or hollow uniform spheres of masses mi and m 2 respectively 
which do not intersect are attracted to each other as if they were particles of the same 
mass situated at their respective geometric centers. 

Since the potential corresponding to 

F = -^^ri (24) 



CHAP. 5] CENTRAL FORCES AND PLANETARY MOTION 121 

is V = -Q™»* {25) 

r 
it is also possible to find the attraction between objects by first finding the potential and 
then using F = -y7. See Problems 5.26-5.33. 

MOTION IN AN INVERSE SQUARE FORCE FIELD 

As we have seen, the planets revolve in elliptical orbits about the sun which is at one 
focus of the ellipse. In a similar manner, satellites (natural or man-made) may revolve around 
planets in elliptical orbits. However, the motion of an object in an inverse square field of 
attraction need not always be elliptical but may be parabolic or hyperbolic. In such cases 
the object, such as a comet or meteorite, would enter the solar system and then leave but 
never return again. 

The following simple condition in terms of the total energy E determines the path of 
an object. 

(i) if E < the path is an ellipse 

(ii) if E = the path is a parabola 

(iii) if E > the path is a hyperbola 

Other conditions in terms of the speed of the object are also available. See Problem 5.37. 

In this chapter we assume the sun to be fixed and the planets do not affect each other. 
Similarly in the motion of satellites around a planet such as the earth, for example, we 
assume the planet fixed and that the sun and all other planets have no effect. 

Although such assumption is correct as a first approximation, the influence of other 
planets may have to be taken into account for more accurate purposes. The problems of 
dealing with the motions of two, three, etc., objects under their mutual attractions are often 
called the two body problem, three body problem, etc. 



Solved Problems 

CENTRAL FORCES AND IMPORTANT PROPERTIES 

5.1. Prove that if a particle moves in a central force field, then its path must be a plane 
curve. 

Let F = f(r) r x be the central force field. Then 

r X F = f(r) r X r x = (1) 

since r 1 is a unit vector in the direction of the position vector r. Since F = mdv/dt, this can be 
written 

r X dx/dt = (2) 

or |(rXv) = (3) 

Integrating, we find r X v = h (-4) 

where h is a constant vector. Multiplying both sides of (4) by r • , 

r«h = (5) 

using the fact that r • (r X v) = (r X r) • v = 0. Thus r is perpendicular to the constant vector h, 
and so the motion takes place in a plane. We shall assume that this plane is taken to be the 
xy plane whose origin is at the center of force. 



122 



CENTRAL FORCES AND PLANETARY MOTION 



[CHAP. 5 



5.2. Prove that for a particle moving in a central force field the angular momentum is 
conserved. 

From equation (4) of Problem 5.1, we have 

r X v = h 
where h is a constant vector. Then multiplying by mass m, 

w(r X v) = mh (1) 

Since the left side of (1) is the angular momentum, it follows that the angular momentum is 
conserved, i.e. is always constant in magnitude and direction. 

EQUATIONS OF MOTION FOR A PARTICLE IN A CENTRAL FIELD 

5.3. Write the equations of motion for a particle in a central field. 

By Problem 5.1 the motion of the particle takes place in a plane. Choose this plane to be 
the xy plane and the coordinates describing the position of the particle at any time t to be 
polar coordinates (r, e). Using Problem 1.49, page 27, we have 

(mass) (acceleration) = net force 

m{(r — re 2 )r 1 + {r'o + 2r'e)0i} = f( r ) r i (■*) 

Thus the required equations of motion are given by 

m(r - re 2 ) = f{r) (2) 

m(r m e + 2r'e) - (8) 

5.4. Show that r 2 6 = h, a constant. 



Method 1. Equation (3) of Problem 5.3 can be written 

m 



m(r e + 2re) 



r 



(r 2 e + 2rre) 



m d_ 
~r di 



(r 2 e) 



Thus 



dt 



(r 2 e) — and so 



r 2 



(1) 



where h is a constant. 

Method 2. By Problem 1.49, page 27, the velocity in polar coordinates is 

v = frj + rbei 

Then from equation (4) of Problem 5.1 

h = r X v = f(r X r x ) + ro(r X oj = r 2 tfk (2) 

since r X r t = and r X $ x — rk where k is the unit vector in a direction perpendicular to the 
plane of motion [the xy plane], i.e. in the direction r X v. Using h = fik in (2), we see that 
r 2 e — h. 



5.5. Prove that r 2 & = 2A where A is the time rate 
at which area is swept out by the position 
vector r. 

Suppose that in time At the particle moves from 
M to N [see Fig. 5-8]. The area AA swept out by the 
position vector in this time is approximately half the 
area of a parallelogram with sides r and Ar or [see 
Problem 1.18, page 15] 



AA = A|rXAr| 



Dividing by Ai and letting 
AA 



lim 

At-»0 At 



lim - 

At-tO 2 



A«-*0, 

Ar I 
At I 



r X 



|r X v| 




Fig. 5-8 



CHAP. 5] CENTRAL FORCES AND PLANETARY MOTION 123 

i.e., A = i|rXv| = \r 2 

using the result in Problem 5.4. Thus r 2 o - 2A, as required. The vector quantity 

A = Ak = i(rXv) = ^(r 2 *)k 
is often called the areal velocity. 

5.6. Prove that for a particle moving in a central force field the areal velocity is constant. 

By Problem 5.4, r 2 = h = a constant. Then the areal velocity is 
A = ±r 2 6k — \hk = ^h, a constant vector 

The result is often stated as follows: If a particle moves in a central force field with O as 
center, then the radius vector drawn from O to the particle sweeps out equal areas in equal 
times. This result is sometimes called the law of areas. 

5.7. Show by means of the substitution r = 1/u that the differential equation for the 
path of the particle in a central field is 

d 2 u , f(l/u) 

+ u — - 



do 2 mh 2 u 2 

From Problem 5.4 or equation (S) of Problem 5.3, we have 

r 2 o = h or = h/r 2 = hu 2 (1) 

Substituting into equation (2) of Problem 5.3, we find 

m(f- h 2 /r 3 ) = f(r) (2) 

Now if r — 1/u, we have 



dr _ dr cfo _ h_ dr_ _ _vdu 
dt ~ de dt ~ r 2 do do 



r = -tt = -^^r = -z— = -*~ (5) 



dr d ( ,du\ d ( , du\ do _ t,2..2 d2u //\ 

r = Tt = dt\- h Te) = de\r h Te)d-t ~ " W ^ (4) 

From this we see that (2) can be written 

m(-h 2 u 2 d 2 u/do 2 - h 2 u?) = f(l/u) (5) 

or, as required, ^ + u = " ^f W 



POTENTIAL ENERGY AND CONSERVATION OF ENERGY 
FOR CENTRAL FORCE FIELDS 

5.8. (a) Prove that a central force field is conservative and (b) find the corresponding 
potential energy of a particle in this field. 

Method 1. 

If we can find the potential or potential energy, then we will have also incidentally proved 
that the field is conservative. Now if the potential V exists, it must be such that 

F-rfr = -dV (1) 

where F — f(r) r x is the central force. We have 

F • dr = f(r) r t • dr = f(r)- • dr = f(r) dr 



since r • dr — r dr. 

Since we can determine V such that 



-dV = f(r)dr 



124 CENTRAL FORCES AND PLANETARY MOTION [CHAP. 5 

for example, V = - J f(r) dr (2) 

it follows that the field is conservative and that (2) represents the potential or potential energy. 

Method 2. 

We can show that V X F = directly, but this method is tedious although straightforward. 

5.9. Write the conservation of energy for a particle of mass m in a central force field. 

Method 1. The velocity of a particle expressed in polar coordinates is [Problem 1.49, page 27] 

v = rr t + reOi s0 tnat y2 = v * v = r2 "*" r2 * 2 

Then the principle of conservation of energy can be expressed as 

imt) 2 + V = E or im(rHrV) - J f(r)dr - E 

where £ is a constant. 

Method 2. The equations of motion for a particle in a central field are, by Problem 5.3, 

m(r — re 2 ) — f(r) (1) 

m(r 6 + 2re) = (2) 

Multiply equation (1) by r, equation (2) by re and add to obtain 

m(r r + r 2 o + rre 2 ) = f(r)r (3) 



This can be written 



lm^(r 2 + r 2 'e 2 ) = ~ f f(r) dr U) 



Then integrating both sides, we obtain 

i m(r 2 + r 2 'e 2 ) - J f(r) dr = E (5) 



5.10. Show that the differential equation describing the motion of a particle in a central 
field can be written as 

™ h2 ^\ 2 + 1 *]-ff(r)dr = E 



2r 4 



l\dej 

From Problem 5.9 we have by the conservation of energy, 

|m(r 2 + r 2 e 2 ) - J f(r) dr = E (1) 



_ dr _ dr do _ dr • 
We also have r ~ dt ~ Tedt ~ ~d6° 

Substituting {2) into (1), we find 



(5) , + ">-J'«"* = E ° r l£[(£) 2+r2 ]-J' w * = E 



since e — h/r 2 . 



5.11. (a) If u=l/r, prove that v 2 = f 2 + W 2 = ^ 2 {(d%/^) 2 + u 2 }. 

(b) Use (a) to prove that the conservation of energy equation becomes 

(du/dOf + v? = 2(E-V)/mh 2 



CHAP. 5] CENTRAL FORCES AND PLANETARY MOTION 125 

(a) From equations (i) and (3) of Problem 5.7 we have 8 = hu 2 , r = —hdu/de. Thus 

v 2 = ' r 2 + r 2^2 = hHdu/de) 2 + (l/u 2 )(hu 2 ) 2 = h 2 {(du/de) 2 + u 2 } 

(b) From the conservation of energy [Problem 5.9] and part (a), 

x mv 2 = i m (f2 + r ' e 2) = e -V or (du/de) 2 + u 2 = 2(E - V)/mh 2 



DETERMINATION OF ORBIT FROM CENTRAL FORCE, 
OR CENTRAL FORCE FROM ORBIT 

5.12. Show that the position of the particle as a function of time t can be determined 
from the equations 

t = J* [G(r)]-"*dr, t = | f r 2 d6 

where G(r) = — + — f f(r) dr - -^^ 

v ' m m J JK ' 2m 2 r 2 

Placing $ — h/r 2 in the equation for conservation of energy of Problem 5.9, 

%m(r 2 + h 2 /r 2 ) - f f(r) dr = E 

or r 2 = — + - f f(r) dr -% - G(r) 

m m J JS ' r 2 ' 

Then assuming the positive square root, we have 

dr/dt = y/Gir) 

and so separating the variables and integrating, we find 

t = f [G(r)]-v*dr 
The second equation follows by writing e — h/r 2 as dt — r 2 de/h and integrating. 



5.13. Show that if the law of central force is denned by 

/(r) = -Kir 2 , K>0 
i.e. an inverse square law of attraction, then the path of the particle is a conic. 

Method 1. 

In this case f(l/u) = —Ku 2 . Substituting into the differential equation of motion in Problem 5.7, 
we find 

d 2 u/de 2 + u = K/mh 2 (l) 

This equation has the general solution 

u = A cos 8 4- B sin + K/mh 2 (2) 

or using Problem 4.2, page 92, 

u = K/mh 2 + C cos {o — <p) {$) 

_ 1 

1,e " r K/mh 2 + C cos (e - 0) W 

It is always possible to choose the axes so that <j> = 0, in which case we have 



K/mh 2 + Ccoss 



126 CENTRAL FORCES AND PLANETARY MOTION [CHAP. 5 



This has the general form of the conic [see Problem 5.16] 

P = 1 

T 1 + e COS 6 1/p + (e/p) COS 



(6) 



Then comparing (5) and (6) we see that 

1/p = tf/mft 2 , e/p = C (7) 

or p = m/i 2 /K, e = mh 2 C/K (8) 

Method 2. Since /(r) = -K/r 2 , we have 

V = - £ f(r) dr = -K/r + c x (9) 

where c x is a constant. If we assume that V -» as r -> °°, then Cj = and so 

V = -ff/r (10) 

Using Problem 5.10, page 124, we find 



mh 2 /dr\ 2 



2r 4 



dej +r2 



E + - (11) 

r 



from which To = ±r \^^ ( } 

By separating variables and integrating [see Problem 5.66] we find the solution (5) where C is 
expressed in terms of the energy E. 



5.14. (a) Obtain the constant C of Problem 5.13 in terms of the total energy E and (b) thus 
show that the conic is an ellipse, parabola or hyperbola according as E < 0, E = 0, 
E > respectively. 

Method 1. 

(a) The potential energy is 

v = - f f(r) dr = f (K/r2) dr = -K/r = -Ku (1) 

where we use u = \/r and choose the constant of integration so that lim V = 0. Now from 
equation (5) of Problem 5.13, 

u - 1/r = K/mh 2 + C cos e (2) 

Thus from Problem 5.11(6) together with (1), we have 

/ K „ V 2E , 2K ( K . n 

(Cs^e^ + i^+C cose) = -^ + ^p(^p + C cos. 



K 2 2E \ K 2 2lT 

c2 = ^p + rf or c = V^ + ^ w 

(6) Using the value of C in part (a), the equation of the conic becomes 



or 

assuming C > 0. 



{■♦V 

Comparing this with (-4) of Problem 5.16, we see that the eccentricity is 



1 K L , , 2ffmft 2 



= ^1+^ (4) 



#2 



From this we see that the conic is an ellipse if £7 < [but greater than -K»/2mfc»}, a parabola 
if E = and a hyperbola if E > 0, since in such cases e < 1, e-1 and e>l respectively. 

Method 2. The value of C can also be obtained as in the second method of Problem 5.13. 



CHAP. 5] 



CENTRAL FORCES AND PLANETARY MOTION 



127 



5.15. Under the influence of a central force at point 0, a particle moves in a circular orbit 
which passes through O. Find the law of force. 

Method 1. 

In polar coordinates the equation of a circle of radius a 
passing through O is [see Fig. 5-9] 

r = 2a cos 9 

Then since u = 1/r = (sec 8)/2a, we have 

du _ sec e tan e 
de ~ 2a 

d 2 u _ (sec fl)(sec 2 e) + (sec 6 tan fl)(tan e) 
de 2 ~~ 2a 

sec 3 9 + sec 9 tan 2 9 



2a 




Thus by Problem 5.7, 



cPu 



f(l/u) = -mh 2 u 2 ( ^-f + u j = -mh 2 u 2 



Fig. 5-9 

/ sec 3 9 + sec 8 tan 2 e + sec e 



mh 2 u 2 
2a 

-8mh 2 a 2 u 5 



{sec 3 + sec (tan 2 + 1)} = 

,. . 8m h 2 a 2 
/(»") = IS— 



2a 

mh 2 u 2 
2a 



2 sec 3 e 



Thus the force is one of attraction varying inversely as the fifth power of the distance from O. 
Method 2. Using r — 2a cos e in equation (16), page 118, we have 

4amh 2 SaHnh 2 



mh 2 2 

— -7- \ —2a cos e — ^ (—2a sin 0) 2 — 2a cos 9 

r 4 2a cos 9 



r 4 cos 9 



CONIC SECTIONS. ELLIPSE, PARABOLA AND HYPERBOLA 
5.16. Derive equation (18), page 118, for a conic section. 

Referring to Fig. 5-3, page 118, by definition of a conic section we have for any point P on it, 

rid = e or d = r/e (1) 

Corresponding to the particular point Q, we have 

p/D = e or p = eD (2) 



But 



t r 

D — d + r cos 9 = — V r cos 9 = — (1 + e cos 8) 



Then from (2) and (3), we have on eliminating D, 

p = r(l + e cos 9) or r 



V 



1 + e COS 



(5) 



(4) 



The equation is a circle if e = 0, an ellipse if < e < 1, a parabola if e = 1 and a hyperbola 
if e > 1. 



5.17. Derive equation (19), page 118, for an ellipse. 

Referring to Fig. 5-4, page 119, we see that when 9 = 0, r = OV and when 9 = v, 
r = OU. Thus using equation (4) of Problem 5.16, 



128 CENTRAL FORCES AND PLANETARY MOTION [CHAP. 5 

OV = p/(l + e), OU = p/(l-e) (-0 

But since 2a is the length of the major axis, 

OV + OU = 2a or p/(l + e) + p/(l - e) = 2a (*) 

from which V = a(l - e 2 ) (3) 

Thus the equation of the ellipse is 



o(l - e 2 ) 

1 + e COS 9 



U) 



5.18. Prove that in Fig. 5-4, page 119, (a) OV = a(l - e ), (b) OU = a(l + e). 

(a) From Problem 5.17, equation (5) and the first equation of (1), 

OV = ^- = 4^ = «( 1 " e ) (i) 

1 + e 1 + e 

(b) From Problem 5.17, equation (3) and the second equation of (1), 

ou = ^- = ^r^ = a < 1 + e > (a) 

1 — e 1—6 

5.19. Prove that c = a £ where c is the distance from the center to the focus of the ellipse. 
a is the length of the semi-major axis and e is the eccentricity. 

From Fig. 5-4, page 119, we have c = CO = CV - OV - a-a(l-e) = ae. 
An analogous result holds for the hyperbola [see Prdblem 5.73(c), page 139]. 

5.20. If a and c are as in Problem 5.19 and b is the length of the semi-minor axis, prove 
that (a) c = \Za 2 - b\ (b) b = a^/l - A 

(a) From Fig. 5-4, page 119, and the definition of an ellipse, we have 

6 = QV = CZ__™ = «__^ or ys = ^=^ (i) 

6 VE VE VE e 

Also since the eccentricity is the distance from O to W divided by the distance from W 
to the directrix AB [which is equal to CE], we have 

OW/CE - e 
or, using (1) and the result of Problem 5.19, 

OW = eCE = e(CV + VE) = e[a + (a-c)/e] = ea + a-c = a 



Then (OW) 2 = (OC) 2 + (CW) 2 or a 2 = b 2 + c 2 , i.e. c = V^ 2 - & 2 - 
(6) From Problem 5.19 and part (a), a 2 = & 2 + a 2 e 2 or 6 = ay/1- e 2 . 



KEPLER'S LAWS OF PLANETARY MOTION AND 
NEWTON'S UNIVERSAL LAW OF GRAVITATION 
5.21. Prove that if a planet is to revolve around the sun in an elliptical path with the sun 

at a focus [Kepler's first law], then the central force necessary varies inversely as 

the square of the distance of the planet from the sun. 

If the path is an ellipse with the sun at a focus, then calling r the distance from the sun, 

we have by Problem 5.16, 



CHAP. 5] CENTRAL FORCES AND PLANETARY MOTION 129 

* X ^costf (1) 



1 + e cos o r p p 

where e < 1. Then the central force is given as in Problem 5.7 by 

f(l/u) = -mh 2 u 2 (d 2 u/de 2 + u) = ^mh 2 u 2 /p (2) 

on substituting the value of u in (1). From (2) we have on replacing u by 1/r, 

f(r) = -mh 2 /pr 2 = -Kir 2 (S) 

5.22. Discuss the connection of Newton's universal law of gravitation with Problem 5.21. 

Historically, Newton arrived at the inverse square law of force for planets by using Kepler's 
first law and the method of Problem 5.21. He was then led to the idea that perhaps all objects 
of the universe were attracted to each other with a force which was inversely proportional to the 
square of the distance r between them and directly proportional to the product of their masses. 
This led to the fundamental postulate 

_ GMm , Y v 

where G is the universal gravitational constant. Equivalently, the law of force (3) of Problem 5.21 
is the same as (1) where 

K = GMm (2) 

5.23. Prove Kepler's third law: The squares of the periods of the various planets are 
proportional to the cubes of their corresponding semi-major axes. 

If a and b are the lengths of the semi-major and semi-minor axes, then the area of the 
ellipse is nab. Since the areal velocity has magnitude h/2 [Problem 5.6], the time taken to sweep 
over area vab, i.e. the period, is 

p _ vab _ 2vab ,^\ 

h/2 ~ h K ' 

Now by Problem 5.17 equation (3), Problem 5.20(6), and Problem 5.13 equation (8), we have 

6 = aVl ~ <?, p = a(l-e 2 ) = mhVK (2) 

Then from (1) and (2) we find 

P = 2irw>i 2 a^ 2 IK^ 2 or P 2 = 4ir 2 maVK 

Thus the squares of the periods are proportional to the cubes of the semi-major axes. 

5.24. Prove that GM = gR 2 . 

On the earth's surface, i.e. r — R where R is the radius, the force of attraction of the earth 
on an object of mass m is equal to the weight mg of the object. Thus if M is the mass of the 

GMm/R 2 = mg or GM = gR 2 

5.25. Calculate the mass of the earth. 

From Problem 5.24, GM = gR 2 or M - gR 2 /G. Taking the radius of the earth as 
6.38 X 10 8 cm, g - 980 cm/sec 2 and G — 6.67 X 10~ 8 cgs units, we find M = 5.98 X 10 27 gra = 
1.32 X 10 25 lb. 



ATTRACTION OF OBJECTS 

5.26. Find the force of attraction of a thin uniform rod of length 2a on a particle of 
mass m placed at a distance b from its midpoint. 



130 



CENTRAL FORCES AND PLANETARY MOTION 



[CHAP. 5 



Choose the x axis along the rod and the y axis 
perpendicular to the rod and passing through its 
center O, as shown in Fig. 5-10. Let a be the mass 
per unit length of the rod. The force of attraction 
dF between an element of mass o dx of the rod and 
m is, by Newton's universal law of gravitation, 



dF 



Gmadx 
x 2 + b 2 
Gmox dx 



(sin e i — cos e j) 
Gmvb dx 




(X 2 + 62)3/2 ( X 2 + 62)3/2 



Fig. 5-10 



since from Fig. 5-10, sin e - x/y/x 2 + b 2 , cos $ - b/y/x 2 + b 2 . Then the total force of attraction is 
_, _ . C a Gmox dx _ . C a Gmabdx 

~ l J ( x 2 +ft 2)3/2 3 J 



= 



(x 2 + b 2 )^ 2 
_. C a Gmabdx 

- 2j X 



(x 2 + ft 2 ) 3 ' 2 



(x 2 + b 2 )*' 2 



Xa 



dx 



(X 2 + 6 2 ) 3 /2 



Let x = b tanfl in this integral. Then when x — 0, e — 0; and when x = a, 6 — tan -1 (a/6). 
Thus the integral becomes 



F = 



-2Gmabj I 



tan 1 (a/b) b se( , 2 Q d$ 



2Gmcra 



(b 2 sec 2 <s>) 3/ 2 



o - ' by/a 2 +b 2 

Since the mass of the rod is M = 2aa, this can also be written as 

^, GMm 
F = 3 

by/a 2 +b 2 

Thus we see that the force of attraction is directed from m to the center of the rod 
and of magnitude 2Gmaa/by/a 2 + b 2 or GMmfby/a 2 + 6 2 . 



5.27. A mass m lies on the perpendicular through the center of a uniform thin circular 
plate of radius a and at distance b from the center. Find the force of attraction 
between the plate and the mass m. 

Method 1. 

Let n be a unit vector drawn from point P where m 
is located to the center O of the plate. Subdivide the 
circular plate into circular rings [such as ABC in 
Fig. 5-11] of radius r and thickness dr. If a is the mass 
per unit area, then the mass of the ring is a{2wrdr). 
S ince all points of the ring are at the same distance 
y/r 2 + b 2 from P, the force of attraction of the ring on 
m will be 

,_ _ Ga{2vr dr)m 
r 2 + 6 2 



cos <f> n 



Go 2irr dr mb 



(1) 



( r 2 4. £,2)3/2 " 

where we have used the fact that due to symmetry the 
resultant force of attraction is in the direction n. By 
integrating over all rings from r = to r = a, we 
find that the total attraction is 



F = 2vGomb 



-r 



r dr 



( r 2 -j. &2)3/2 



b \Vr 2 + fe 2 




Fig. 5-11 



(2) 



To evaluate the integral, let r 2 + 6 2 



so that r dr = u du. Then since u = 6 when r = 



and u = yja 2 + b 2 when r — a, the result is 



CHAP. 5] 



CENTRAL FORCES AND PLANETARY MOTION 



131 



IwGomb n I 



v^+b* udu 



2irGom n ( 1 — 



u ° \ y/a 2 +b 2 , 

If we let a be the value of $ when r = a, this can be written 

F = lirGom n (1 - cos a) (3) 

Thus the force is directed from m to the center O of the plate and has magnitude 2irG<rm(l — cos a). 

Method 2. 

The method of double integration can also be used. In such case the element of area at A is 
rdrde where 6 is the angle measured from a line [taken as the x axis] in the plane of the 
circular plate and passing through the center O. Then we have as in equation (1), 

Ga(r dr do)mb 



d¥ 



and by integrating over the circular plate 



( r 2 -(- J,2)3/2 



F = Gamb 



■/* S 



r=0 ^0 = 



r dr de 
(r 2 + 6 2 ) 3/2 



Xa 



2irr dr 
(r 2 + 6 2 )3/ 2 



2TrGom n (1 — cos a) 



5.28. A uniform plate has its boundary con- 
sisting of two concentric half circles of 
inner and outer radii a and b respec- 
tively, as shown in Fig. 5-12. Find the 
force of attraction of the plate on a 
mass m located at the center 0. 

It is convenient to use polar coordinates 
(r, e). The element of area of the plate [shaded 
in Fig. 5-12] is dA = r dr de, and the mass is 
ardrde. Then the force of attraction between 
dA and O is 

_ _ G{ar dr de)m 




Fig. 5-12 



(cos e i + sin e j) 



Thus the total force of attraction is 



XC G{ardrde)m . . . . n .. 

I — - — -£ — — (cos e i + sin e j) 



)=0 r=a 

Gam In 



W X=o 



(cos o i + sine j) de = 2Gaw In ( - ) j 



Since M — o(±vb 2 — ^va 2 ), we have a = 2Mh(b 2 — a 2 ) and the force can be written 

AGMm . fb\ . 
* = ^2-^2) ^ (^-j, 

The method of single integration can also be used by dividing the region between r = a and 
r — b into circular rings as in Problem 5.27. 



5.29. Find the force of attraction of a thin spherical shell of radius a on a particle P of 
mass m at a distance r > a from its center. 

Let O be the center of the sphere. Subdivide the surface of the sphere into circular elements 
such as ABCDA of Fig. 5-13 below by using parallel planes perpendicular to OP. 

The area of the surface element ABCDA as seen from Fig. 5-13 is 

2^{a sin e)(a de) = 2va 2 sin e de 

since the radius is a sin e [so that the perimeter is 2ir(a sin »)] and the thickness is ade. Then 
if a is the mass per unit area, the mass of ABCDA is 2rra 2 a sin e de. 



132 



CENTRAL FORCES AND PLANETARY MOTION 



[CHAP. 5 



Since all points of ABCDA are at the same dis- 
tance w = AP from P, the force of attraction of 
the element ABCDA on m is 



7 „ G(2va 2 a sin 6 de)m 

d¥ = s cos <p n 

in* 



(1) 



where we have used the fact that from symmetry the 
net force will be in the direction of the unit vector n 
from P toward O. Now from Fig. 5-13, 



PE 
AP 



PO - EO 
AP 



r — a cos e 

w 



(2) 



Using (2) in (1) together with the fact that by the 
cosine law 

w 2 = a 2 + r 2 — 2ar cos B (3) 

we find 
dF 



G{2ira 2 a sin e de)m{r — a cos e) 




ade 



(a 2 + r 2 — 2ar cos 0) 3/2 
Then the total force is 

F = 2TrGa 2 am 



Fig. 5-13 



"J. 



(r — a cos e) sin e 
(a 2 + r 2 — 2ar cos 0) 3/2 



de 



(•4) 



We can evaluate the integral by using the variable w given by (3) in place of 6. When 



= 0, 



2ar + r 2 = (r — a) 2 so that w = r — a if r > a. Also when 6 = ir, w 2 — 



a 2 + 2ar + r 2 = (r + a) 2 so that w = r + a. In addition, we have 

2w dw = 2ar sin 6 de 

' a 2 -f- r 2 _ w 2\ «)2 



a COS 



2ar 



a 2 + r 2 



Then (4) becomes 



F = 



irGaam n 



x: 



i + 



dw — 



2r 

&TrGa 2 om n 



r 2 



5.30. Work Problem 5.29 if r < a. 

In this case the force is also given by (4) of Problem 5.29. However, in evaluating the 
integral we note that on making the substitution (3) of Problem 5.29 that e = yields w 2 = (a — r) 2 
or w — a — r if r < a. Then the result (4) of Problem 5.29 becomes 



F = 



wGaom 



Jo 



1 - 



dw = 



Thus there will be no force of attraction of a spherical shell on any mass placed inside. This 
means that in such case a particle will be in equilibrium inside of the shell. 



5.31. Prove that the force of attraction in Problem 5.29 is the same as if all the mass of 
the spherical shell were concentrated at its center. 

The mass of the shell is M = AttcPo. Thus the force is F = (GMm/r 2 )n, which proves the 
required result. 



5.32. (a) Find the force of attraction of a solid uniform sphere on a mass m placed outside 
of it and (b) prove that the force is the same as if all the mass were concentrated 
at its center. 

(a) We can subdivide the solid sphere into thin concentric spherical shells. If p is the distance 
of any of these shells from the center and dp is the thickness, then by Problem 5.29 the 
force of attraction of this shell on the mass m is 



CHAP. 5] 



CENTRAL FORCES AND PLANETARY MOTION 



133 



_ G<x(4t7 P 2 d P )m 
d¥ — • — s n 



(1) 



where a is the mass per unit volume. Then the total force obtained by integrating from 
r = to r = a is 

AirGomn f a „ . 

^ dp 



F = 



Xa 
> 2 



G{% va s )am n 



(*) 



(6) Since the mass of the sphere is M — ^waPo, (2) can be written as F = (GMm/r 2 )n, which 
shows that the force of attraction is the same as if all the mass were concentrated at 
the center. 

We can also use triple integration to obtain this result [see Problem 5.130]. 



5.33. Derive the result of Problems 5.29 and 5.30 by first finding the potential due to the 
mass distribution. 

The potential dV due to the element ABCDA is 

G{2tto, 2 o sin e do)m __ G{2va 2 a sin e de)m 



dV 



Then the total potential is 



— 2irGa 2 am 



V« 2 + r 2 — 2ar cos 9 



o V a2 + r 2 — 2ar cos 
_ 2irGaam ^/ ( „ + y)2 -,/(„.- r)2 } 



If r > a this yields 
If r < a it yields 



4:irGa 2 cnn 



GMm 



r r 

V = —kwGaom 



Then if r > a the force is 

F = -VV = -V 
and if r < a the force is 



GMm \ 



GMm 



,2 l l 



F = -VV = -V(-4s-Gaam) = 
in agreement with Problems 5.29 and 5.30. 



MISCELLANEOUS PROBLEMS 

5.34. An object is projected vertically upward from the earth's surface with initial 
speed Vo. Neglecting air resistance, (a) find the speed at a distance H above the 
earth's surface and (6) the smallest velocity of projection needed in order that the 
object never return. 

(a) Let r denote the radial distance of the object at time t 
from the center of the earth, which we assume is fixed 
[see Pig. 5-14]. If M is the mass of the earth and R is 
its radius, then by Newton's universal law of gravitation 
and Problem 5.29, the force between m and M is 



F = 



GMm 



(1) 



where rj is a unit vector directed radially outward from 
the earth's center in the direction of motion of the object. 

If v is the speed at time t, we have by Newton's sec- 
ond law, 




Fig. 5-14 



134 CENTRAL FORCES AND PLANETARY MOTION [CHAP. 5 



tt) 

Since the object starts from the earth's surface with speed v , we have v = v when r = R 
so that <?! = vjj/2 - GM/.R. Then (4) becomes 





GMm 
r 2 r i 




dv 

Tt 


This can be written as 








dv dr _ 
dr dt 


GM 

r 2 


or 


dv 
dr 


Then by integrating, we find 


v 2 /2 = 


GM/r + c t 



2GM (i-i) + ^ 

ve the earth' 



Thus when the object is at height H above the earth's surface, i.e. r = R + H, 

2GMH 



(5) 



R(R + H) 



, 2 2GMH 
V = ^ V °-R(R + H) 

Using Problem 5.24, this can be written 



v = ^jvl 



!~2 2gRH 

V = \<~R+H 



(6) 



(b) As H -* oo, the limiting speed (6) becomes 

vX ~ 2GM/R or v^ 2 , - 2gR (7) 

since lim /p „ = 1. The minimum initial speed occurs where (7) is zero or where 

H-t» (it + /3) 



v = V2GM/R = y/2gR (8) 

This minimum speed is called the escape speed and the corresponding velocity is called the 
escape velocity from the earth's surface. 



5.35. Show that the magnitude of the escape velocity of an object from the earth's surface 
is about 7 mi/sec. 

From equation (*) of Problem 5.34, v = y/2gR. Taking g = 32 ft/sec 2 and R = 4000 mi, 
we find v — 6.96 mi/sec. 



5.36. Prove, by using vector methods primarily, that the path of a planet around the sun 
is an ellipse with the sun at one focus. 

Since the force F between the planet and sun is 

F _ w dv _ GMm 

¥ ~ m dt ~ r 2 ri {1) 

, dv GM ,-. 

we have -=r = x-r t (2) 

dt r 2 

Also, by Problem 5.1, equation (4), we have 

r X v = h (3) 

dv av^ (£y 

Now since r = rr., v = -37 — r—r- + -r-r,. Thus from (S), 
at at at 

f dr, (j r \ dr, 

h = rX v = „ lX (,— + -„) = ^r.x-J W) 



CHAP. 5] CENTRAL FORCES AND PLANETARY MOTION 135 

From (2), 

dv GM f dr i 

[f *A drA or, 

= - GM {("-^>-<'.-")-«i = GM ^r 

using equation (4) above and equation (7), page 5. 

But since h is a constant vector, -jr X h = — (v X h) so that 

at at 

d ^ r i 

|(vxh) = GM-± 

Integrating, v X h = GM r t + c 

from which 

r • (v X h) = GM t • r 1 + r • c = GMr + rr x • c = GMr + re cos e 

where c is an arbitrary constant vector having magnitude c, and e is the angle between c and r x . 
Since r • (v X h) = (r X v) • h = h • h = h 2 [see Problem 1.72(a), page 27], 

h 2 = GMr + re cos 6 
h 2 h 2 /GM 



and so 



GM + c cos e 1 + (c/GM) cos e 



which is the equation of a conic. Since the only conic which is a closed curve is an ellipse, the 
required result is proved. 

5.37. Prove that the speed v of a particle moving in an elliptical path in an inverse square 
field is given by 



m \r a 
where a is the semi-major axis. 

By (8) of Problem 5.13, (4) of Problem 5.14 and (3) of Problem 5.17, we have 

mh 2 n 2 . / 2Emh 2 \ ... 

V = -g^ = a(l-e 2 ) = a I K2~J (*) 

from which E - -K/2a (2) 

Thus by the conservation of energy we have, using V = —Kir, 

or v 2 = -(---) (3) 

m \r a I 



Kf2 +h) (*) 



We can similarly show that for a hyperbola, 



m \r a 

while for a parabola [which corresponds to letting a -* °° in either (3) or (4)] , 

v 2 = 2K/mr 



5.38. An artificial (man-made) satellite revolves about the earth at height H above the 
surface. Determine the (a) orbital speed and (b) orbital period so that a man in the 
satellite will be in a state of weightlessness. 

(a) Assume that the earth is spherical and has radius R. Weightlessness will result when the 
centrifugal force [equal and opposite to the centripetal force, i.e. the force due to the cen- 



136 



CENTRAL FORCES AND PLANETARY MOTION 



[CHAP. 5 



tripetal acceleration] acting on the man due to rotation of the satellite just balances his 
attraction to the earth. Then if v is the orbital speed, 

mv l _ GMm _ gR*m _ R 

R + H ~ (R + H)* ~ (R + H)* or v ° ~ R + H 

If H is small compared with R, this is y/Rg approximately. 



y/(R + H)g 



(b) 



Orbital speed 



distance traveled in one revolution 
time for one revolution, or period 



Thus 

Then from part (a) 



_ 2tt(R + H) 



_ 2u(R + H) _ 
r — — ^ 7 

«0 



R + H 
R 



: )V 1 ? 



If H is small compared with R, this is 2iry/R/g approximately. 



5.39. Calculate the (a) orbital speed and (b) period in Problem 5.38 assuming that the 
height H above the earth's surface is small compared with the earth's radius. 

Taking the earth's radius as 4000 miles and g = 32 ft/sec 2 , we find (a) v = y/Rg - 
4.92 mi/sec and (b) P = 2-n-^R/g - 1.42 hr = 85 minutes, approximately. 



5.40. Find the force of attraction of a solid sphere of radius a on a particle of mass m at 
a distance b < a from its center. 

By Problem 5.30 the force of attraction of any spherical 
shell containing m in its interior [such as the spherical shell 
shown dashed in Fig. 5-15] is zero. 

Thus the force of attraction on m is the force due to a 
sphere of radius b < a with center at O. If a is the mass 
per unit volume, the force of attraction is 

G(f7r&3)<rm/&2 = (^Gam)b 

Thus the force varies as the distance 6 from the mass to the 

center. Fig. 5-15 




Supplementary Problems 



CENTRAL FORCES AND EQUATIONS OF MOTION 

5.41. Indicate which of the following central force fields are attractive toward origin O and which 
are repulsive from O. (a) F = -4^; (6) F = Kxjyfr, K > 0; (c) F = r(r- ljr^r 2 + 1); 
(d) F = sin wr r v 

Ans. (a) attractive; (6) repulsive; (<;) attractive if < r < 1, repulsive if r > 1; (d) repulsive for 
2n < r < 2n + 1, attractive for 2n+l<r<2n + 2 where n — 0, 1, 2, 3 . . . . 

5.42. Prove that in rectangular coordinates the magnitude of the areal velocity is ^{xy — yx). 

5.43. Give an example of a force field directed toward a fixed point which is not a central force field. 



CHAP. 5] CENTRAL FORCES AND PLANETARY MOTION 137 

5.44. Derive equation (7), page 117. 

5.45. If a particle moves in a circular orbit under the influence of a central force at its center, prove that 
its speed around the orbit must be constant. 

5.46. A particle of mass m moves in a force field defined by F = -Kr^r 3 . If it starts on the positive 
x axis at distance a away from the origin and moves with speed v in direction making angle 
a with the positive x axis, prove that the differential equation for the radial position r of the 
particle at any time Ms „ „ 

d2 r _ (K — ma 2 v\ sin 2 a) 

dt 2 mr 3 

5.47. (a) Show that the differential equation for the orbit in Problem 5.46 is given in terms of u = 1/r by 

+ (1 — y)u — where y — 



do 2 ' ma 2 v 2 sin 2 a 

(b) Solve the differential equation in (a) and interpret physically. 

5.48. A particle is to move under the influence of a central force field so that its orbital speed is 
always constant and equal to v . Determine all possible orbits. 

POTENTIAL ENERGY AND CONSERVATION OF ENERGY 

5.49. Find the potential energy or potential corresponding to the central force fields defined by 
(a) F = -Krjr*, (b) F = {air 2 + 0/r»)r lf (c) F = Krr u (d) F = r x /yfF, (e) F = sin a-r r^ 

Ans. (a) -K/2r 2 , (b) a/r + p/2r 2 , (c) \Kr 2 , (d) 2ypr, (e) (cos»r)/ir 

5.50. (a) Find the potential energy for a particle which moves in the force field F = —Krjr 2 . (b) How 
much work is done by the force field in (a) in moving the particle from a point on the circle 
r = a > to another point on the circle r = 6 > 0? Does the work depend on the path? Explain. 
Ans. (a) -Kir, (6) K(a-b)/ab 

5.51. Work Problem 5.50 for the force field F = -Krjr. Ans. (a) -K\nr, (6) -K In (a/6) 

5.52. A particle of mass m moves in a central force field defined by F = —Krjr*. (a) Write an equation 
for the conservation of energy. (6) Prove that if E is the total energy supplied to the particle, 
then its speed is given by v = yjKlmr 2 + 2E/m. 

5.53. A particle moves in a central force field defined by F = —Kr 2 r v It starts from rest at a point 
o n the circle r = a. (a) Prove that when it reaches the circle r = b its speed will be 
V2#(a 3 - 6 3 )/3w and that (6) the speed will be independent of the path. 

5.54. A particle of mass m moves in a central force field F = Kxjr n where K and n are constants. 
It starts from rest at r = a and arrives at r = with finite speed v . (a) Prove that we must 
have n < 1 and K > 0. (b) Prove that v = y/2Ka i - n /m{n- 1). (c) Discuss the physical sig- 
nificance of the results in (a). 

5.55. By differentiating both sides of equation (IS), page 117, obtain equation (6). 

DETERMINATION OF ORBIT FROM CENTRAL FORCE OR 
CENTRAL FORCE FROM ORBIT 

5.56. A particle of mass m moves in a central force field given in magnitude by f(r) = —Kr where 
K is a positive constant. If the particle starts at r — a, e = with a speed v in a direction 
perpendicular to the x axis, determine its orbit. What type of curve is described? 

5.57. (a) Work Problem 5.56 if the speed is v in a direction making angle a with the positive * axis. 
(6) Discuss the cases a — 0, a = ir and give the physical significance. 



138 CENTRAL FORCES AND PLANETARY MOTION [CHAP. 5 

5.58. A particle moving in a central force field located at r = describes the spiral r = e~ . Prove 
that the magnitude of the force is inversely proportional to r 3 . 

5.59. Find the central force necessary to make a particle 
describe the lemniscate r 2 = a 2 cos 2d [see Fig. 5-16]. 
Ans. A force proportional to r~ 7 . r 2 = a*cos2» 

5.60. Obtain the orbit for the particle of Problem 5.46 and 
describe physically. 

5.61. Prove that the orbits r — e~ e and r = 1/8 are both 
possible for the case of an inverse cube field of force. 
Explain physically how this is possible. Fig. 5-16 

5.62. (a) Show that if the law of force is given by 




F = 



r 4 cos e r 2 cos 3 e 

then a particle can move in the circular orbit r — 2a cos 8. (b) What can you conclude about 
the uniqueness of forces when the orbit is specified? (c) Answer part (b) when the forces are 
central forces. 

5.63. (a) What central force at the origin O is needed to make a particle move around O with a speed 
which is inversely proportional to the distance from O. (b) What types of orbits are possible in 
such case? Ans. (a) Inverse cube force. 

5.64. Discuss the motion of a particle moving in a central force field given by F = (a/r 2 + )8/r 3 )r 1 . 

5.65. Prove that there is no central force which will enable a particle to move in a straight line. 

5.66. Complete the integration of equation (12) of Problem 5.13, page 125 and thus arrive at equation 
(5) of the same problem. [Hint. Let r = 1/u.] 

5.67. Suppose that the orbit of a particle moving in a central force field is given by 8 = e(r). Prove 

mh 2 [2e' + re" + r 2 (e') 3 ] 
that the law of force is 5 , ,. 3 where primes denote differentiations with 

respect to r. 

5.68. (a) Use Problem 5.67 to show that if 6 = 1/r, the central force is one of attraction and varies 
inversely as r 3 . (6) Graph the orbit in (a) and explain physically. 

CONIC SECTIONS. ELLIPSE, PARABOLA AND HYPERBOLA 

12 

5.69. The equation of a conic is r = -^—, • Graph the conic, finding (a) the foci, (6) the vertices, 

S + cos 8 

(c) the length of the major axis, (d) the length of the minor axis, (e) the distance from the center 
to the directrix. 

24 

5.70. Work Problem 5.69 for the conic r = 



3 + 5 cos 8 ' 

5.71. Show that the equation of a parabola can be written as r — p sec 2 (8/2). 

5.72. Find an equation for an ellipse which has one focus at the origin, its center at the point (—4, 0), 
and its major axis of length 10. Ans. r = 9/(5 + 4 cos 8) 



CHAP. 5] CENTRAL FORCES AND PLANETARY MOTION 139 

5.73. In Fig. 5-17, SR or TN is called the minor axis of %\ x/' 
the hyperbola and its length is generally denoted >^ // 

by 26. The length of the major axis VU is 2a, \ x r re's 

while the distance between the foci O and O' is 2c \l X \ / / \f 

[i.e. the distance from the center C to a focus O m _\ n ., x y 

or O' is C]. °/ y / /Cx U \ ' 

(a) Prove that c 2 = a 2 + b 2 . /y y _ ^H\ 

(b) Prove that b = aVe 2 — 1 where e is the eccen- // \ 
tricity. >^ x \ 

(c) Prove that c = etc. Compare with results for y ^> 
the ellipse. Fig. 5-17 

5.74. Derive equation (22), page 119, for a hyperbola. 

5.75. In rectangular coordinates the equations for an ellipse and hyperbola in standard form are given by 

* 2 ,y 2 1 , x 2 v 2 

tf + V 2 = 1 and V 2 ~V = x 

respectively, where a and b are the lengths of the semi-major and semi-minor axes. Graph these 
equations, locating vertices, foci and directrices, and explain the relation of these equations to 
equations [19), page 118, and (22), page 119. 

5.76. Using the alternative definitions for an ellipse and hyperbola given on pages 118-119, obtain the 
equations (19) and (22). 

5.77. Prove that the angle between the asymptotes of a hyperbola is 2 cos -1 (1/e). 

KEPLER'S LAWS AND NEWTON'S LAW OF GRAVITATION 

5.78. Assuming that the planet Mars has a period about the sun equal to 687 earth days approximately, 
find the mean distance of Mars from the sun. Take the distance of the earth from the sun as 
93 million miles. Ans. 140 million miles 

5.79. Work Problem 5.78 for (a) Jupiter and (b) Venus which have periods of 4333 earth days and 
225 earth days respectively. Ans. (a) 484 million miles, (b) 67 million miles 

5.80. Suppose that a small spherical planet has a radius of 10 km and a mean density of 5 gm/cm 3 . 

(a) What would be the acceleration due to gravity at its surface? (6) What would a man weigh 
on this planet if he weighed 80 kgwt on earth? 

5.81. If the acceleration due to gravity on the surface of a spherically shaped planet P is g P while its 
mean density and radius are given by a P and R P respectively, prove that g P = %irGR P e P where G 
is the universal gravitational constant. 

5.82. If L, M, T represent the dimensions of length, mass and time, find the dimensions of the universal 
gravitational constant. Ans. L 3 M _1 T~ 2 

5.83. Calculate the mass of the sun using the fact that the earth is approximately 150 X 10 6 kilometers 
from it and makes one complete revolution about it in approximately 365 days. Ans. 2 X 10 30 kg 

5.84. Calculate the force between the sun and the earth if the distance between the earth and the sun is 
taken as 150 X 10 6 kilometers and the masses of the earth and sun are 6 X 10 24 kg and 2 X 10 30 kg 
respectively. Ans. 1.16 X 10 24 newtons 

ATTRACTION OF OBJECTS 

5.85. Find the force of attraction of a thin uniform rod of length a on a mass m outside the rod but on 
the same line as the rod and distance 6 from an end. Ans. GMm/b(a + b) 

5.86. In Problem 5.85 determine where the mass of the rod should be concentrated so as to give the 
same force of attraction. Ans. At a point in the rod a distance y/b(a + b) — 6 from the end 



14 () CENTRAL FORCES AND PLANETARY MOTION [CHAP. 5 

5.87. Find the force of attraction of an infinitely long thin uniform rod on a mass m at distance b 
from it. Arts. Magnitude is 2Gmafb 

5.88. A uniform wire is in the form of an arc of a circle of radius 6 and central angle ^. Prove that 
the force of attraction of the wire on a mass m placed at the center of the circle is given in 
magnitude by 2Gam sin (v fr/2) 

6V ° r * 

where M is the mass of the wire and a is the mass per unit length. Discuss the cases ^ = v/2 
and yp — 7r. 

5.89. In Fig. 5-18, AB is a thin rod of length 2a and m 
is a mass located at point C a distance 6 from the 

rod. Prove that the force of attraction of the rod . 

on m has magnitude ^^ c 

— — sin Ua + (3) D/ 

ao ^\ 

/^ \ 6 \ 

in a direction making an angle with the rod s' \ \ 

given by /^ v \^^ E ^A~ 

_ 1 / cos/3 + cosoA ^ r ~~ ~~' '■■-""" ^ /? 

tan VsiniS-sinay I 2a I 

Discuss the case a = /? and compare with Prob- 
lem 5.26. Fig. 5-18 

5.90. By comparing Problem 5.89 with Problem 5.88, prove that the rod of Problem 5.89 can be 
replaced by a wire in the form of circular arc BEG [shown dashed in Fig. 5-18] which has its 
center at C and is tangent to the rod at E. Prove that the direction of the attraction is toward 
the midpoint of this arc. 

5.91. A hemisphere of mass M and radius a has a particle of mass m located at its center. Find the 
force of attraction if (a) the hemisphere is a thin shell, (6) the hemisphere is solid. 

Ana. (a) GMm/2a 2 , (b) 36 Mm/2a 2 

5.92. Work Problem 5.91 if the hemisphere is a shell having outer radius a and inner radius b. 

5.93. Deduce from Kepler's laws that if the force of attraction between sun and planets is given in 
magnitude by ym/r 2 , then y must be independent of the particular planet. 

5 94 A cone has height H and radius a. Prove that the force of attraction on a particle of mass m 

■. ^ §GMm A H \ 

placed at its vertex has magnitude — ^ — 1 1 ~ / 24- W2 / " 

5.95. Find the force of attraction between two non-intersecting spheres. 

5 96 A particle of mass m is placed outside of a uniform solid hemisphere of radius a at a distance a 
on a line perpendicular to the base through its center. Prove that the force of attraction is 
given in magnitude by GMm{\f2 — l)/a 2 . 

5.97. Work (a) Problem 5.26, (b) Problem 5.27, and (c) Problem 5.94 by first finding the potential. 

MISCELLANEOUS PROBLEMS 

5.98. A particle is projected vertically upward from the earth's surface with initial speed v . 

(a) Prove that the maximum height H reached above the earth's surface is H = v 2 R/(2gR - v\). 

(b) Discuss the significance of the case where v\ = 2gR. 

(c) Prove that if H is small, then it is equal to v%/2g very nearly. 



CHAP. 5] CENTRAL FORCES AND PLANETARY MOTION 141 

5.99. (a) Prove that the time taken to reach the maximum height of Problem 5.98 is 



J R + H J [H 

^~2tUr 



~ R + H _JR-H 



2R wo \R + H t 
(b) Prove that if H is very small compared with R, then the time in (a) is very nearly ^2H/g. 

5.100. (a) Prove that if an object is dropped to the earth's surface from a height H, then if air 

resistance is negligible it will hit the earth with a speed v - ^2gRH/(R + H) where R is 
the radius of the earth. 

(b) Calculate the speed in part (a) for the cases where H = 100 miles and H = 10,000 miles 
respectively. Take the radius of the earth as 4000 miles. 

5.101. Find the time taken for the object of Problem 5.100 to reach the earth's surface in each of 
the two cases. 

5.102. What must be the law of force if the speed of a particle in a central force field is to be 
proportional to r~ n where n is a constant? 

5.103. What velocity must a space ship have in order to keep it in an orbit around the earth at a 
distance of (a) 200 miles, (6) 2000 miles above the earth's surface" 



>9 



5.104. An object is thrown upward from the earth's surface with velocity v . Assuming that it returns 
to earth and that air resistance is negligible, find its velocity on returning. 

5.105. (a) What is the work done by a space ship of mass m in moving from a distance a above the 

earth's surface to a distance 6? 

(6) Does the work depend on the path? Explain. Ans. (a) GmM(a — b)/ab 

5.106. (o) Prove that it is possible for a particle to move in a circle of radius a in any central force 

field whose law of force is /(r). 

(b) Suppose the particle of part (a) is displaced slightly from its circular orbit. Prove that 
it will return to the orbit, i.e. the motion is stable, if 

w 4U .l • af ' {a) + 3/(a) > ° 

but is unstable otherwise. 

(c) Illustrate the result in (6) by considering f(r) = 1/r" and deciding for which values of n 
stability can occur. Ans. (c) For n < 3 there is stability. 

5.107. If the moon were suddenly stopped in its orbit, how long would it take to fall to the earth 
assuming that the earth remained at rest? Ans. About 4 days 18 hours 

5.108. If the earth were suddenly stopped in its orbit, how long would it take for it to fall into the sun? 
Ans. About 65 days 

5.109. Work Problem 5.34, page 133, by using energy methods. 

5.110. Find the velocity of escape for an object on the surface of the moon. Use the fact that the 
acceleration due to gravity on the moon's surface is approximately 1/6 that on the earth and 
that the radius of the moon is approximately 1/4 of the earth's radius. Ans. 1.5 mi/sec 

5.111. An object is dropped through a hole bored through the center of the earth. Assuming that the 
resistance to motion is negligible, show that the speed of the particle as it passes through the 
center of the earth is slightly less than 5 mi/sec. 

[Hint. Use Problem 5.40, page 136.] 

5.112. In Problem 5.111 show that the time taken for the object to return is about 85 minutes. 



14 2 CENTRAL FORCES AND PLANETARY MOTION [CHAP. 5 

5.113. Work Problems 5.111 and 5.112 if the hole is straight but does not pass through the center of 
the earth. 

5.114. Discuss the relationship between the results of Problems 5.111 and 5.112 and that of Problem 5.39. 

5.115. How would you explain the fact that the earth has an atmosphere while the moon has none? 

5.116. Prove Theorem 5.1, page 120. 

5.117. Discuss Theorem 5.1 if the spheres intersect. 

5.118. Explain how you could use the result of Problem 5.27 to find the force of attraction of a solid 
sphere on a particle. 

5.119. Find the force of attraction between a uniform circular ring of outer radius a and inner radius 6 
and a mass m located on its axis at a distance 6 from its center. 

5.120. Two space ships move about the earth on the same elliptical path of eccentricity e. If they are 
separated by a small distance D at perigee, prove that at apogee they will be separated by the 
distance D(l - e)/(l + c). 

5.121. (a) Explain how you could calculate the velocity of escape from a planet. (6) Use your method to 
calculate the velocity of escape from Mars. Ans. (6) 5 km/sec, or about 3 mi/sec 

5.122. Work Problem 5.121 for (a) Jupiter, (b) Venus. Ans. (a) about 38 mi/sec, (6) about 6.3 mi/sec 

5.123. Three infinitely long thin uniform rods having the same mass per unit length lie in the same plane 
and form a triangle. Prove that force of attraction on a particle will be zero if and only if the 
particle is located at the intersection of the medians of the triangle. 

5.124. Find the force of attraction between a uniform rod of length a and a sphere of radius b if they 
do not intersect and the line of the rod passes through the center. 

5.125. Work Problem 5.124 if the rod is situated so that a line drawn from the center perpendicular to the 
line of the rod bisects the rod. 

5.126. A satellite of radius a revolves in a circular orbit about a planet of radius b with period P. 
If the shortest distance between their surfaces is c, prove that the mass of the planet is 
4^-2(a + b + c)3/GP 2 . 

5.127. Given that the moon is approximately 240,000 miles from the earth and makes one complete 
revolution about the earth in 27£ days approximately, find the mass of the earth. 

Ans. 6 X 10 24 kg 

5.128. Discuss the relationship of Problem 5.126 with Kepler's third law. 

5.129. Prove that the only central force field F whose divergence is zero is an inverse square force field. 

5.130. Work Problem 5.32, page 132, by using triple integration. 

5.131. A uniform solid right circular cylinder has radius a and height H. A particle of mass m is placed 
on the extended axis of the cylinder so that it is at a distance D from one end. Prove that the 
force of attraction is directed along the axis and given in magnitude by 

^^{H + Va2 + Z)2- Va2 + (D + H)2} 
a 2 H 

5.132. Suppose that the cylinder of Problem 5.131 has a given volume. Prove that the force of attraction 
when the particle is at the center of one end of the cylinder is a maximum when alH = £(9 - y/ll). 

5.133. Work (a) Problem 5.26 and (&) Problem 5.27 assuming an inverse cube law of attraction. 



CHAP. 5] CENTRAL FORCES AND PLANETARY MOTION 143 

5.134. Do the results of Problems 5.29 and 5.30 apply if there is an inverse cube law of attraction? 
Explain. 

5.135. What would be the velocity of escape from the small planet of Problem 5.80? 

5.136. A spherical shell of inner radius a and outer radius 6 has constant density a. Prove that the 
gravitational potential V(r) at distance r from the center is given by 



V(r) = 



2ua(b 2 -a 2 ) r < a 

27r(j(b 2 - lr 2 ) - 4™a 3 /3r a < r < b 
47tct(6 3 — a 3 )/3r r > b 



5.137. If Einstein's theory of relativity is taken into account, the differential equation for the orbit 
of a planet becomes 

d 2 u , K , . 

do 2 mh 2 ' 

where y = 3K/mc 2 , c being the speed of light, (a) Prove that if axes are suitably chosen, then 
the position r of the planet can be determined approximately from 

mh 2 /K , 1 „, L9 

r ■= - — ; where a — 1 — yK/mh 2 

1 + e cos aO 

(b) Use (a) to show that a planet actually moves in an elliptical path but that this ellipse slowly 
rotates in space, the rate of angular rotation being 2iryK/mh 2 . (c) Show that in the case of 
Mercury this rotation amounts to 43 seconds of arc per century. This was actually observed, thus 
offering experimental proof of the validity of the theory of relativity. 

5.138. Find the position of a planet in its orbit around the sun as a function of time t measured from 
where it is furthest from the sun. 

5.139. At apogee of 200 miles from the earth's surface, two space ships in the same elliptical path are 
500 feet apart. How far apart will they be at perigee 150 miles assuming that they drift without 
altering their path in any way? 

5.140. A particle of mass m is located on a perpendicular line through the center of a rectangular plate 
of sides 2a and 26 at a distance D from this center. Prove that the force of attraction of the plate 
on the particle is given in magnitude by 

GMm . _j / ab 

ab Sin \^(a 2 + D 2 )(b 2 + D 2 ) 

5.141. Find the force of attraction of a uniform infinite plate of negligible thickness and density a 
on a particle at distance D from it. Ans. 2iroGm 

5.142. Points where r = are called apsides [singular, apsis], (a) Prove that apsides for a central 
force field with potential V(r) and total energy E are roots of the equation V(r) + h 2 /2r 2 = E. 
(b) Find the apsides corresponding to an inverse square field of force, showing that there are 
two, one or none according as the orbit is an ellipse, hyperbola or parabola. 

5.143. A particle moving in a central force field travels in a path which is the cycloid r = a(l-cose). 
Find the law of force. Ans. Inverse fourth power of r. 

5.144. Set up equations for the motion of a particle in a central force field if it takes place in a medium 
where the resistance is proportional to the instantaneous speed of the particle. 



5.145. A satellite has its largest and smallest orbital speeds given by v max and i; min respectively. Prove 
that the eccentricity of the orbit in which the satellite moves is equal to 



max ' v r 



5.146. Prove that if the satellite of ProbLem 5.145 has a period equal to t, then it moves in an elliptical 
path having major axis whose length is - y/v max v min . 



Chapter 6 



MOVING 
COORDINATE SYSTEMS 



NON-INERTIAL COORDINATE SYSTEMS 

In preceding chapters the coordinate systems used to describe the motions of particles 
were assumed to be inertial [see page 33]. In many instances of practical importance, 
however, this assumption is not warranted. For example, a coordinate system fixed in 
the earth is not an inertial system since the earth itself is rotating in space. Consequently 
if we use this coordinate system to describe the motion of a particle relative to the earth 
we obtain results which may be in error. We are led therefore to consider the motion of 
particles relative to moving coordinate systems. 

ROTATING COORDINATE SYSTEMS 

In Fig. 6-1 let XYZ denote an inertial coordinate 
system with origin O which we shall consider fixed 
in space. Let the coordinate system xyz having the 
same origin O be rotating with respect to the XYZ 
system. 

Consider a vector A which is changing with 
time. To an observer fixed relative to the xyz system 
the time rate of change of A = Aii + A 2 j + A 3 k is 
found to be 



dA 
dt 



dAi 
~df 



i + 



dA, 
dt 



i + 



dAt 
dt 



(1) 



where subscript M indicates the derivative in the 
moving (xyz) system. 

However, the time rate of change of A relative 
to the fixed XYZ system symbolized by the subscript 
F is found to be [see Problem 6.1] 




Fig. 6-1 



dA 
dt 



dA 
dt 



+ o> x A 



(2) 



where a> is called the angular velocity of the xyz system with respect to the XYZ system. 

DERIVATIVE OPERATORS 

Let D F and D M represent time derivative operators in the fixed and moving systems. 
Then we can write the operator equivalence 

D F = D M + oX (3) 

This result is useful in relating higher order time derivatives in the fixed and moving 
systems. See Problem 6.6. 



144 



CHAP. 6] MOVING COORDINATE SYSTEMS 145 

VELOCITY IN A MOVING SYSTEM 

If, in particular, vector A is the position vector r of a particle, then (2) gives 



+ *>Xr a) 



dr _ dr 

dt f ~ dt 

or D F r = D M x + <o X r (5) 

Let us write 

v p|F = dr/dt \ F = D F r = velocity of particle P relative to fixed system 

v p(M = dr/dt \ M = D M r = velocity of particle P relative to moving system 

v mif = • x r = velocity of moving system relative to fixed system. 

Then (-4) or (5) can be written 

v pif = v P|M + v M|F (6) 

ACCELERATION IN A MOVING SYSTEM 

If D 2 F = d 2 /dt 2 \ F and D^ = d 2 /dt 2 | M are second derivative operators with respect to t 
in the fixed and moving systems, then application of (3) yields [see Problem 6.6] 

D 2 F r = D 2 M r + (Z> M <o) X r + 2« X D M r + • X (• x r) (7) 

Let us write 

a P|F = d 2 r/dt 2 \ F = D F r = acceleration of particle P relative to fixed system 

a P|M = d 2 r/dt 2 \ M = D^r = acceleration of particle P relative to moving system 

a Mir = ( D m<*) X r + 2a. X D M r + « X (o, X r) 

= acceleration of moving system relative to fixed system 

Then (7) can be written 



*P|F **P|M 



"*" a M|F (#) 



X r (11) 

M/ 



CORIOLIS AND CENTRIPETAL ACCELERATION 

The last two terms on the right of (7) are called the Coriolis acceleration and centripetal 
acceleration respectively, i.e., 

Coriolis acceleration = 2o» x D M r — 2<o x v M (9) 

Centripetal acceleration = • x («• x r) (10) 

The second term on the right of (7) is sometimes called the linear acceleration, i.e., 

/j 

Linear acceleration = (D^a) x r = (~ 

v M ' \dt 

and Z> M o> is called the angular acceleration. For many cases of practical importance [e.g. in 
the rotation of the earth] <a is constant and D M <a = 0. 

The quantity — <o x (o> x r) is often called the centrifugal acceleration. 

MOTION OF A PARTICLE RELATIVE TO THE EARTH 

Newton's second law is strictly applicable only to inertial systems. However, by using 
(7) we obtain a result valid for non-inertial systems. This has the form 

mD^r = F - m(D M a) X r - 2m(u x D M r) - m«x( B Xr) (12) 

where F is the resultant of all forces acting on the particle as seen by the observer in the 
fixed or inertial system. 



146 



MOVING COORDINATE SYSTEMS 



[CHAP. 6 



In practice we are interested in expressing the equations of motion in terms of quantities 
as determined by an observer fixed on the earth [or other moving system]. In such case 
we may omit the subscript M and write (12) as 
d 2 v 



m 



dt 2 



m(m X r) — 2m(o> x v) — m[» x (© X r)] 



(13) 



For the case of the earth rotating with constant angular o> about its axis, *> = and 
(13) becomes , 2 



m 



dt 2 



2m(<o X v) - m[» X (o> x r)] 



(U) 



CORIOLIS AND CENTRIPETAL FORCE 

Referring to equations (13) or (14) we often use the following terminology 
Coriolis force = 2ra(« x r) = 2ra(o> x v) 

Centripetal force = m[« x (» x r)] 
Centrifugal force = -m[» x (» x r)] 



MOVING COORDINATE SYSTEMS IN GENERAL 

In the above results we assumed that the coordi- 
nate systems xyz and XYZ [see Fig. 6-1] have com- 
mon origin O. In case they do not have a common 
origin, results are easily obtained from those already 
considered. 

Suppose that R is the position vector of origin Q 
relative to origin [see Fig. 6-2]. Then if R and 
R denote the velocity and acceleration of Q relative 
to O, equations (5) and (7) are replaced respectively 
by 



D F r 



= R + ZX„r + <oXr 



= R + 



dr 
dt 



+ <» xr 



(15) 




and 



Fig. 6-2 
Dl* = R + D 2 M r + (D M <o) x r + 2. x D M r + . x („ x r) 

= R +S+ iXr + 2«Xv + «)X( a Xr) 



Similarly equation (14) is replaced by 



m 



^r 
dt 2 



= F - 2m(u X v) - m[<a x(<aX r)] - mR 



(16) 
(17) 



THE FOUCAULT PENDULUM 

Consider a simple pendulum consisting of a long string and heavy bob suspended 
vertically from a f rictionless support. Suppose that the bob is displaced from its equilibrium 
position and is free to rotate in any vertical plane. Then due to the rotation of the earth, 
the plane in which the pendulum swings will gradually precess about a vertical axis. In the 
northern hemisphere this precession is in the clockwise direction if we look down at the 
earth's surface. In the southern hemisphere the precession would be in the counterclock- 
wise direction. 

Such a pendulum used for detecting the earth's rotation was first employed by Foucault 
in 1851 and is called Foucault's pendulum. 



CHAP. 6] 



MOVING COORDINATE SYSTEMS 



147 



Solved Problems 



ROTATING COORDINATE SYSTEMS 

6.1. An observer stationed at a point which is fixed relative to an xyz coordinate system 
with origin O [see Fig. 6-1, page 144] observes a vector A = Ad + A 2 j + Ask. and 

calculates its time derivative to be —fri + ~rfj~i + ~dT^- Later, he finds that he 
and his coordinate system are actually rotating with respect to an XYZ coordinate 
system taken as fixed in space and having origin also at 0. He asks, "What would 
be the time derivative of A for an observer who is fixed relative to the XYZ coordi- 
nate system?" 



If 



dA 
dt 



j dA 
and —rr 

f dt 



denote respectively the time derivatives of A relative to the 



fixed and moving systems, show that there exists a vector quantity <*> such that 



dA 
dt 



dA 
dt 



+ a X A 



To the fixed observer the unit vectors i, j,k actually change with time. Hence such an 
observer would compute the time derivative as 



dA 
dt 



i.e., 



dA x 
dt 

dAI 
dt \f 



dA 



2 . 



dt 



dA 3 ^ 
dt 



i + ^i + -^k + A^ + A 2 f t + A ZTt 



di 
dt 



d\ 



dk 



dAl 
dt \m 



, di . d) a dk 

+ ^1^7 + A 2^ + A 3 — 



dt 



dt 



dt 



(1) 
(2) 



Since i is a unit vector, di/dt is perpendicular to i and must therefore lie in the plane 
of j and k. Then 

di/dt = « J + a 2 k (S) 



Similarly, 



dj/dt = a 3 k + a 4 i 
dk/dt = a 5 i + a 6 j 

. dj ,di . _ 



U) 



>dy 



di 



From i«j = 0, differentiation yields i * jT+ ji* j = 0- But i» -3i—a 4 from (4) and -Tz'j = a l 



from (3). Thus a 4 = — a v 



dt dt 



dt 



dt 



dk , di 
dt dt 



Similarly from i«k = 0, i»-^- + ^»k = and a 5 = —a 2 ; from j»k = 0, J'-^t+Tt* k = 
and a 6 = — a 3 . Then 



.rfk+dl.i, - 
dt dt 



di/dt = aj + a 2 k, dj/dt = a 3 k — a t i, dk/dt = —a 2 i — atf 



It follows that 

. di 



^dt 
which can be written as 



+ A 2 ^ + A s -^ = (-«iA 2 - a 2 A 3 )i + («iA l -a s A s )j + (a 2 A! + a 3 A 2 )k (6) 



dt 



i 


j 


k 


«3 


_ «2 


<*i 


Ai 


A 2 


A 3 



Then if we choose a 3 = « lf — a 2 = <o 2 , a t = w 3 this determinant becomes 

i J k 



«! W 2 <0 3 



= <■> X A 



where <a = u 1 i + « 2 j + w 3 k. 



148 



MOVING COORDINATE SYSTEMS 



[CHAP. 6 



From (2) and (6) we find, as required, 

dA dA 

dt j? dt 



+ w X A 



The vector quantity o» is the angular velocity of the moving system relative to the fixed system. 

6.2. Let D F and D M be symbolic time derivative operators in the fixed and moving systems 
respectively. Demonstrate the operator equivalence 



By definition D F A 



dA 
dt 

DmA = ~di 



D F = D M + «x 



= derivative in fixed system 



= derivative in moving system 



Then from Problem 6.1, 

D F A = D M A + »XA = (D M + co X)A 

which shows the equivalence of the operators D F = D M + o> X . 

6.3. Prove that the angular acceleration is the same in both XYZ and xyz coordinate 

systems. 

Let A = <o in Problem 6.1. Then 



da 

~dt 



da 
~dl 



+ to X 



da 
~di 



Since dm/dt is the angular acceleration, the required statement is proved. 

VELOCITY AND ACCELERATION IN MOVING SYSTEMS 

6.4. Determine the velocity of a moving particle as seen by the two observers in 
Problem 6.1. 

Replacing A by the position vector r of the particle, we have 



dr 
dt 



dr 
dt 



+ o, X r 



(1) 



If r is expressed in terms of the unit vectors i, j,k of the moving coordinate system, then the 
velocity of the particle relative to this system is, on dropping the subscript M, 



dr 
dt 



dx . dy^ . dz , 
dt di 3 dt 



and the velocity of the particle relative to the fixed system is from (1) 



dr 
~di 



dr . v 
dt + wXr 



(2) 



(3) 



The velocity (3) is sometimes called the true velocity, while (2) is the apparent velocity. 



6.5. An xyz coordinate system is rotating with respect to an XYZ coordinate system 
having the same origin and assumed to be fixed in space [i.e. it is an inertial system]. 
The angular velocity of the xyz system relative to the XYZ system is given by 
a) = 2ti - t 2 j + (2t + 4)k where t is the time. The position vector of a particle at 
time t as observed in the xyz system is given by r = (t 2 + l)i - 6ij + 4£ 3 k. Find 
(a) the apparent velocity and (b) the true velocity at time t = 1. 



CHAP. 6] MOVING COORDINATE SYSTEMS 149 

(a) The apparent velocity at any time t is 

dr/dt = 2ti - 6j + 12t 2 k 
At time t = 1 this is 2i - 6j + 12k. 

(6) The true velocity at any time t is 

dr/dt I.Xr = (2ti - 6j + 12t%) + [2ti - t 2 ) + (2t + 4)k] X [(t 2 + l)i - 6£j + 4£%] 
At time £ = 1 this is 



2i - 6j + 12k + 



i J k 

2-16 
2-6 4 



34i - 2j + 2k 



6.6. Determine the acceleration of a moving particle as seen by the two observers in 
Problem 6.1. 

The acceleration of the particle as seen by the observer in the fixed XYZ system is 
D F x — D F {D F r). Using the operator equivalence established in Problem 6.2, we have 

D F (D F r) = D F (D M r + *> X r) 

= (D M + «» X )(D M r + a. x r) 
= D M {D M r + » X r) + u X (D M r + .Xr) 
= D^r + D M (a> X r) + o> X D M r + «* X («• X r) 
or since D M (<* X r) = (D M m) X r + » X (D M r), 

Dlr - DmT + (D MW ) X r + 2.X (D M r) + • X (• X r) (1) 

If r is the position vector expressed in terms of i, j,k of the moving coordinate system, then 
the acceleration of the particle relative to this system is, on dropping the subscript M, 

<ffr _ d?x. d 2 y. d*z 

dt 2 dt 2 * + dt 2 * dt 2 (g) 

The acceleration of the particle relative to the fixed system is given from (1) as 

d 2 r I (Pr , d«„ , „ .. /dr 

|f 

The acceleration (3) is sometimes called the true acceleration, while (2) is the apparent acceleration 



dt 2 \ F ^ + ^ Xr + 2wX U^ + ttX(wXr) w 



6.7. Find (a) the apparent acceleration and (b) the true acceleration of the particle in 
Problem 6.5. 

(a) The apparent acceleration at any time t is 

d 2 r d fdr\ d 



dt 2 ~ dt\dtj ~ eft (2«- 61 + 12^) - 2i + 24*k 
At time t - 1 this is 2i + 24k. 
(b) The true acceleration at any time t is 

At time t — 1 this equals 

2i + 24k + (4i - 2j + 12k) X (2i - 6j + 12k) 

+ (2i - 2j + 2k) X (2i - 6j + 4k) 

+ (2i - j + 6k) X {(2i - j + 6k) X (2i - 6j + 4k)} 

= 2i + 24k + (48i - 24j - 20k) + (4i - 4j - 8k) + (-141 + 212j + 40k) 
= 40i + 184j + 36k 



150 MOVING COORDINATE SYSTEMS [CHAP. 6 

CORIOLIS AND CENTRIPETAL ACCELERATION 

6.8. Referring to Problem 6.5, find (a) the Coriolis acceleration, (b) the centripetal 
acceleration and (c) their magnitudes at time t = 1. 

(a) From Problem 6.5 we have, 

Coriolis acceleration = 2« X dr/dt = (4i - 2j + 12k) X (2i - 6 j + 12k) 
= 48i - 24j - 20k 

(b) From Problem 6.5 we have, 

Centripetal acceleration = a. X (• X r) = (2i - j + 6k) X (32i + 4j - 10k) 
= -14i + 212j + 40k 

(c) From parts (a) and (6) we have 

Magnitude of Coriolis acceleration = V(48) 2 + (-24)2 + (_ 2 o)2 = 4\/205 
Magnitude of centripetal acceleration = V(-14) 2 + (212)2 + ( 40 )2 = 2\/ll,685 



MOTION OF A PARTICLE RELATIVE TO THE EARTH 

6.9. (a) Express Newton's second law for the motion of a particle relative to an XYZ 
coordinate system fixed in space (inertial system). (6) Use (a) to find an equation of 
motion for the particle relative to an xyz system having the same origin as the XYZ 
system but rotating with respect to it. 

(a) If m is the mass of the particle (assumed constant), d 2 r/dt 2 \ F its acceleration in the fixed 
system and F the resultant of all forces acting on the particle as viewed in the fixed system, 
then Newton's second law states that 

d 2 r 



m 



dt 2 \F 



= F (1) 



(b) Using subscript M to denote quantities as viewed in the moving system, we have from 
Problem 6.6, 



dt 2 



d?r I .... . „ w dt 



§-§ +iXr+2.X^I +.X(.Xr) («) 

dt 2 \m dt \m 

Substituting this into (1), we find the required equation 

m^§ = F - m(iXr) - 2w(. xfjl ) - m[. X (. X r)] (*) 

dt 2 m \ dt \m j 

We can drop the subscript M provided it is clear that all quantities except F are as 
determined by an observer in the moving system. The quantity F, it must be emphasized, is the 
resultant force as observed in the fixed or inertial system. If we do remove the subscript M 
and write dr/dt = v, then (3) can be written 

= F - w(i X r) - 2m(» X v) — m[» X (*> X r)] (4) 



d 2 v 



6.10. Calculate the angular speed of the earth about its axis. 

Since the earth makes one revolution [2tt radians] about its axis in approximately 24 hours = 
86,400 sec, the angular speed is 

w = 86^400 = 7.27 X 10-5 rad/sec 

The actual time for one revolution is closer to 86,164 sec and the angular speed 7.29 X lO" 5 rad/sec. 



CHAP. 6] 



MOVING COORDINATE SYSTEMS 



151 



MOVING COORDINATE SYSTEMS IN GENERAL 

6.11. Work Problem 6.4 if the origins of the XYZ 
and xyz systems do not coincide. 

Let R be the position vector of origin Q of the 
xyz system relative to origin O of the fixed (or inertial) 
XYZ system [see Fig. 6-3]. The velocity of the par- 
ticle P relative to the moving system is, as before, 

dr\ _ dv 

dt \m dt 



dx. . dy . . dz . 
-i + "rrj + — k 



dt 



dt 



dt 



(1) 



Now the position vector of P relative to O is 
P = R + r and thus the velocity of P as viewed in the 
XYZ system is 



dp 

dt ~ 



di (R + r) \ F " H\f + di 
dv 



= * + dt + » Xr ^ 



using equation (3) of Problem 6.4. Note that R is the 
velocity of Q with respect to O. If R = this re- 
duces to the result of Problem 6.4. 




Fig. 6-3 



6.12. Work Problem 6.6 if the origins of the XYZ and xyz systems do not coincide. 

Referring to Fig. 6-3, the acceleration of the particle P relative to the moving system is, 
as before, 

= dfr &x i+ d*yd?z 

m dt 2 dt 2 ^ dt 2 ' dt 2 K1) 



d?r 
dt 2 



Since the position vector of P relative to O is p = R + r, the acceleration of P as viewed in the 
XYZ system is 



d 2 p 
dt 2 



J^+'i 



dm, 

dt 2 



d 2 r 

F dt 2 



= R + S + ^Xr + 2.X*+«X( tt Xr) 



dt 2 



dt 



dt 



(2) 



using equation (S) of Problem 6.6. Note that R is the acceleration of Q with respect to O. If 
R = this reduces to the result of Problem 6.6. 



6.13. Work Problem 6.9 if the origins of the XYZ and xyz systems do not coincide. 

(a) The position vector of the particle relative to the fixed (XYZ) system is p. Then the 
required equation of motion is 



d?P 
dt 2 



(1) 



(6) Using the result (2) of Problem 6.12 in (1), we obtain 
d 2 r 



dt 2 



= F 



mR — m(i X r) — 2m(<* X v) — m[» X (o> X r)] 



where F is the force acting on m as viewed in the inertial system and where v = r. 



6.14. Find the equation of motion of a particle relative to an observer on the earth's 
surface. 



152 



MOVING COORDINATE SYSTEMS 



[CHAP. 6 



We assume the earth to be a sphere with center 
at [Fig. 6-4] rotating about the Z axis with 
angular velocity » = wK. We also use the fact that 
the effect of the earth's rotation around the sun is 
negligible, so that the XYZ system can be taken as 
an inertial system. 

Then we can use equation (2) of Problem 6.12. 
For the case of the earth, we have 

m = (1) 

R = o> X (o> X R) (2) 

F = -2»E, ,„ 

the first equation arising from the fact that the ro- 
tation of the earth about its axis proceeds with con- 
stant angular velocity, the second arising from the 
fact that the acceleration of origin Q relative to O 
is the centripetal acceleration, and the third arising 
from Newton's law of gravitation. Using these in 
(2) of Problem 6.12 yields the required equation, 




Fig. 6-4 



dt* 



GM 



p — a) X (*» X R) — 2(<* X v) - « X (» X r) 



(4) 



assuming that other forces acting on m [such as air resistance, etc.] are neglected. 



We can define 



GM . 

g-p — tt X («* X R) 



as the acceleration due to gravity, so that (4) becomes 

|^ = g - 2(» X v) - » X (• X r) 



(5) 



(6) 



Near the earth's surface the last term in (6) can be neglected, so that to a high degree of 
approximation , 






= g - 2(» X v) 



(7) 



In practice we choose g as constant in magnitude although it varies slightly over the earth's 
surface. If other external forces act, we must add them to the right side of equations (6) or (7). 



6.15. Show that if the particle of Problem 6.14 moves near the earth's surface, then the 
equations of motion are given by 

x = 2«> cos A y 

y = — 2(w cos A x + m sin A z) 

z = —g + 2o>sinXy 

where the angle A is the colatitude [see Fig. 6-4] and 90° - A is the latitude. 

From Fig. 6-4 we have 

K = (K-i)i + (K-j)j + (K-k)k 

= (—sin X)i + Oj + (cos X)k = -sin X i + cos X k 

an( j s0 «» = mK = —w sin X i + w cos X k 



CHAP. 6] MOVING COORDINATE SYSTEMS 153 

Then » X v = » X (xi + yj + zk) 

i j k 

— <o sin X w cos \ 

• • • 

x y z 

= (— w cos X y)\ + (u cos \ x + u sin X z)j — (« sin X y)k 
Thus from equation (7) of Problem 6.14 we have 

= — gk. + 2« cos X y i — 2(« cos \ x + a sin X z)j + 2<o sin X 2/ k 

Equating corresponding coefficients of i, j,k on both sides of this equation, we find, as required, 

x = 2w cos X y (1) 

y = — 2(u cos X x + <o sin X z) (2) 

z — — g + 2u sin X y (3) 

6.16. An object of mass m initially at rest is dropped to the earth's surface from a height 
which is small compared with the earth's radius. Assuming that the angular speed 
of the earth about its axis is a constant «>, prove that after time t the object is 
deflected east of the vertical by the amount %<»gt 3 sin A. 

Method 1. 

We assume that the object is located on the z axis at * = 0, y = 0, z — h [see Fig. 6-4]. From 
equations (i) and (2) of Problem 6.15 we have on integrating, 

x = 2(o cos X y + c it V — — 2(w cos X x + <o sin X z) + c 2 

Since at t = 0, x = 0, y = 0, x = 0, y = 0, z = ft we have c x = 0, c 2 = 2u sin X ft. Thus 

x = 2<o cos X j/, y = — 2(w cos X x + w sin X z) + 2a sin X ft (i) 

Then (5) of Problem 6.15 becomes 

'z = — g + 2w sin X y — —g — 4u 2 sin X [cos X a; + sin X (z — h)] 

But since the terms on the right involving u 2 are very small compared with — g we can neglect them 
and write z = —g. Integration yields z = —gt + c 3 . Since 1 = at t = 0, we have c 3 = or 

z = -gt (2) 

Using equation (2) and the first equation of (1) in equation (2) of Problem 6.15 we find 
j/ = (—2(0 cos X)(2w cos X 2/) + (— 2u sin X)(—gt) 
= — 4w 2 cos 2 X 1/ + 2w sin X gt 

Then neglecting the first term, we have y = 2« sin X gt. Integrating, 

y — wg sin X t 2 + c 4 

Since y = at £ = 0, we have c 4 = and y — ag sin X t 2 . Integrating again, 

2/ = ^ug sin X t 3 + c 5 

Then since y = at £ = 0, c 5 = so that, as required, 

y - ^ug sin X t 3 (3) 

Method 2. 

Integrating equations (!), (£) and (£) of Problem 6.15, we have 

x = 2w cos X 2/ + c x 

y = — 2(w cos X x + (o sin X z) + c 2 

z = —gt + 2w sin \y + c z 



154 



MOVING COORDINATE SYSTEMS 



[CHAP. 6 



Using the fact that at t = 0, 
c 2 = 2o>h sin X, c 3 = 0. Thus 



x = y — z = and 



0; y = 0, 



h, we have e x = 0, 



2w cos X i/ 

— 2(<o cos X » + w sin X 2) + 2uh sin X 

— gt + 2w sin X y 



Integrating these we find, using the above conditions, 

2w cos X I y du 



J" 



y — 2aht sin X — 2w cos XI x du — 2w sin X 






r 



2 dw 



— h — igt 2 + 2w sin X 






y du 



(5) 
(6) 



Since the unknowns are under the integral sign, these equations are called integral equations. 
We shall use a method called the method of successive approximations or method of iteration to 
obtain a solution to any desired accuracy. The method consists of using a first guess for x, y, z 
under the integral signs in (4), (5) and (6) to obtain a better guess. As a first guess we can try 
x — 0, y = 0, z = under the integral signs. Then we find as a second guess 

x = 0, 1/ = 2wfti sin X, z - h — ±gt 2 

Substituting these in (4), (5) and (6) and neglecting terms involving <o 2 , we find the third guess 



x - 0, y = 2cofct sin X — 2<o sin X(7ii — ^g&) 



— ^ug& sin X, 



z — h 



&* 



Using these in (4), (5) and (6) and again neglecting terms involving w 2 , we find the fourth guess 



0, 



[ugt 3 sin X, 



= h 



i9t 2 



Since this fourth guess is identical with the third guess, these results are accurate up to terms 
involving w 2 , and no further guesses need be taken. It is thus seen that the deflection is 
y = j^ugt 3 sin X, as required. 

6.17. Referring to Problem 6.16, show that an object dropped from height h above the 
earth's surface hits the earth at a point east of the vertical at a distance 
%oih sin A yjlhlg. 

From (2) of Problem 6.16 we have on integrating, z=_-±gt 2 + c. Since z-h at t - 0, c-h 
and z-h- \g&. Then at z = 0, h = ±gt 2 or t = V^ft/fl'. Substituting this value of t into (5) 
of Problem 6.16, we find the required distance. 



THE FOUCAULT PENDULUM 

6.18. Derive an equation of motion for a simple 
pendulum, taking into account the earth's 
rotation about its axis. 

Choose the xyz coordinate system of Fig. 6-5. 
Suppose that the origin O is the equilibrium position 
of the bob B, A is the point of suspension and the 
length of string AB is I. If the tension in the string 
is T, then we have 

T = (T-i)i + (T-j)j + (T-k)k 

= T cos a i + T cos p j + T cos y k 
' x\ . „, / y\ . , mil — z 



= -T 



Ti l 



j + T 



I 



(1) 



Since the net force acting on B is T + mg, the equa- 
tion of motion of B is given by [see Problem 6.14] 

m^J- = T + mg - 2m(« X v) - m<* X (• X r) (*) 




Fig. 6-5 



CHAP. 6] MOVING COORDINATE SYSTEMS 155 

If we neglect the last term in (2), put g = — gk and use (1), then (2) can be written in component 
form as 

m x = —T(x/l) + 2muy cos \ (S) 

my = —T(y/l) — 2mu(x cos \ + z sin X) (4) 

m*z = T(l — z)/l — mg + 2mJy sin X (5) 

6.19. By assuming that the bob of the simple pendulum in Problem 6.18 undergoes small 
oscillations about the equilibrium position so that its motion can be assumed to take 
place in a horizontal plane, simplify the equations of motion. 

Making the assumption that the motion of the bob takes place in a horizontal plane amounts 
to assuming that z and z are zero. For small vibrations (l — z)/l is very nearly equal to one. 
Then equation (5) of Problem 6.18 yields 

= T — mg + 2muy sin X 
or T = mg — 2muy sin X (1) 

Substituting (1) into equations (3) and (4) of Problem 6.18 and simplifying, we obtain 

gx . 2axy sinX . „ . . ,„•> 

x = — ^j- H —j h 2wj/cosX (2) 

gy , 2ayy sin X „ • . /ff v 

y = —^j- + — ^y 2 <oa; cosX (3) 

These differential equations are non-linear because of the presence of the terms involving xy 
and yy. However, these terms are negligible compared with the others since «, x and y are 
small. Upon neglecting them we obtain the linear differential equations 

x = —gx/l + 2uy cos X (-4) 

y = —gy/l — 2(o* cos X (5) 

6.20. Solve the equations of motion of the pendulum obtained in Problem 6.19, assuming 
suitable initial conditions. 

Suppose that initially the bob is in the yz plane and is given a displacement from the z axis 
of magnitude A > 0, after which it is released. Then the initial conditions are 

x = 0, x = 0, y = A, y = at t = (1) 

To find the solution of equations (-4) and (5) of Problem 6.19, it is convenient to place 

K 2 = g/l, a = a cos X (2) 

so that they become x — —K 2 x + 2ay (8) 

y = -K 2 y - 2ax U) 

It is also convenient to use complex numbers. Multiplying equation (4) by i and adding to (3), 
we find 

x + iy = -K 2 (x + iy) + 2 a (y-ix) = -K 2 (x + iy) - 2ia(x + iy) 

Then calling u = x + iy, this can be written 

u = -K 2 u - 2iau or u + 2iau + K 2 u = (5) 

If u = Cey* where C and y are constants, this becomes 

y 2 + 2iay + K 2 = 

so that V = (-2i« ± V / -4a 2 - 4K 2 )/2 = -ia ± Wa 2 + K 2 (6) 

Now since a 2 = w 2 cos 2 X is small compared to K 2 = g/l, we can write 

y = — ia ± iK (7) 



156 MOVING COORDINATE SYSTEMS [CHAP. 6 

Then solutions of the equation are (allowing for complex coefficients) 

(C 1 + iC 2 )e- Ka - K > t and (C 3 + iC 4 )e-«« + K>t 
and the general solution is 

u = (C^iCJe-*"-™ +■ (C 3 + tC 4 )e-«« + K)t (5) 

where d> C 2 , C 3 , C 4 are assumed real. Using Euler's formulas 

e i0 = costf + ism 6, e~ ie = cos e — t sin (9) 

and the fact that u = x + iy, (8) can be written 

x + iy = (C x + t'C 2 ){cos (a - K)t - i sin (a - K)£} + (C 3 + iC 4 ){cos (« + #)£ - i sin (a + K)t) 
Equating real and imaginary parts, we find 

x = d cos (« - #)* + C 2 sin (« - K)t + C 3 cos (a + K)t + C 4 sin (a + X)t (10) 

y = -dam (a -K)t + C 2 cos (a -K)t - C 3 sin (a + K)t + C 4 cos (a + K)t (11) 

Using the initial condition x = at * = 0, we find from (10) that d + C 3 = or 
C 3 = -d. Similarly, using x = at t = 0, we find from (10) that 

X _ a \ /Vff/l- wcosX 



c * " C »U + J " C2 Vx^77+ w cosx, 

Now since a cos X is small compared with -\fgjl, we have, to a high degree of approximation, 
C 4 = C 2 . 

Thus equations (10) and (ii) become 

x = d cos (« - X)t + C 2 sin (a - K)t - d cos ( a + K ) 1 + C 2 sin (« + K)t (12) 

y = -d sin (a - K)t + C 2 cos (a - K)t + d sin (a + K)« + C 2 cos (a + X)t (15) 

Using the initial condition y = 0, (1*) yields d = 0- Similarly using y = A at t = 0, we 
find C 2 = -|A. Thus (1«) and (15) become 

x = \A sin (a - K)t + |A sin (a + K)t 
y = \A cos (a - K)« + \A cos (a + #)* 

or x = A cos Kt sm at "1 

2/ = A COSK* COS at J 

i e., a; = A cosy/gilt sin(ucosXt) 1 

i- (■*£) 

j/ = A cos V^ t cos (w cos H) J 

6.21. Give a physical interpretation to the solution (15) of Problem 6.20. 

In vector form, (15) can be written 

r = x\ + 2/j = A cos V 'gilt n 
w here n = i sin (« cos X)t + j cos (w cos \)t 

is a unit vector. 

The period of cos yfgllt [namely, 2ttV^] is very small compared with the period of n [namely, 
2tt/((ocosX)]. It follows that n is a very slowly turning vector. Thus physically the pendulum 
oscillates in a plane through the z axis which is slowly rotating (or precessing) about the z axis. 

Now at t = 0, n = j and the bob is at y = A. After a time t = 2^/(4 w cosX), for example, 
n = l\/2i + J-V^j so that the rotation of the plane is proceeding in the clockwise direction as 
viewed from 2 above the earth's surface in the northern hemisphere [where cosX>0]. In the 
southern hemisphere the rotation of the plane is counterclockwise. 

The rotation of the plane was observed by Foucault in 1851 and served to provide laboratory 
evidence of the rotation of the earth about its axis. 



CHAP. 6] 



MOVING COORDINATE SYSTEMS 



157 




MISCELLANEOUS PROBLEMS 

6.22. The vertical rod AB of Fig. 6-6 is rotating with 
constant angular velocity <■>. A light inextensible 
string of length I has one end attached at point O 
of the rod while the other end P of the string has a 
mass m attached. Find (a) the tension in the string 
and (b) the angle which string OP makes with the 
vertical when equilibrium conditions prevail. 

Choose unit vectors i and k perpendicular and parallel 
respectively to the rod and rotating with it. The unit vector 
j can be chosen perpendicular to the plane of i and k. Let 

r = I sin e i — I cos e k 
be the position vector of m with respect to O. 

Three forces act on particle m 
(i) The weight, mg = — mgk 

(ii) The centrifugal force, 

-m{a X (» X r)} = -m{[tok] X ([wk] X [I sin e i - I cos B k])} 
= — m{[wk] X (al sin e j)} = ma 2 l sin i 

(iii) The tension, T = —T sin e i + T cos 6 k 

When the particle is in equilibrium, the resultant of all these forces is zero. Then 

—mgk. + muH sin 6 i — T sin e i + T cos k = 

i.e., (mu 2 l sin d — T sin $)i + (T cos 6 — mg)k = 

or tow 2 / sin 6 — T sin - (1) 

T cos 6 — mg = (2) 

Solving (1) and (2) simultaneously, we find (a) T = mu 2 l, (6) o = cos -1 (g/u 2 l). 

Since the string OP with mass m at P describe the surface of a cone the system is sometimes 
called a conical pendulum. 



Fig. 6-6 



6.23. A rod AOB [Fig. 6-7] rotates in a vertical plane [the yz plane] about a horizontal 
axis through O perpendicular to this plane [the x axis] with constant angular 
velocity w. Assuming no frictional forces, determine the motion of a particle P of 
mass m which is constrained to move along the rod. An equivalent problem exists 
when the rod A OB is replaced by a thin hollow tube inside which the particle can move. 




Fig. 6-7 

At time t let r be the position vector of the particle and e the angle made by the rod with 
the y axis. Choose unit vectors j and k in the y and z directions respectively and unit vector 
i = j X k. Let rj be a unit vector in the direction r and e 1 a unit vector in the direction of 
increasing e. 



158 MOVING COORDINATE SYSTEMS [CHAP. 6 

There are three forces acting on P: 
(i) The weight, rag = — mgk — —mg sin r x — mg cos 9 X 

(ii) The centrifugal force, 

-ra[o» X («* X r)] = -ra[wi X («i X rr t )] 

= — ra[coi(wi • rrj.) — rr 1 (wi • wi)] 

= — ra[0 — wVrj] = ra-uVr! 

(iii) The reaction force N = N8 t of the rod which is perpendicular to the rod since there are 
no frictional or resistance forces. 



Then by Newton's second law, 
d 2 x 
dt 2 



d 2 x 
ra-^To — — ra^k + ra<o 2 rr x + 2V0J 



or m-7r^r 1 = —mg sine r t — mg cos 0j + mu 2 rr x + iVOj 

= (ma 2 r — mg sin 0)r x + (N — mg cos ff)^ 

It follows that N = mg cos and 

d 2 r/d« 2 = u 2 r - g sin (i) 

Since 6 = a, a constant, we have 6 = wt if we assume = at t = 0. Then (i) becomes 

d 2 r/d£ 2 - w 2 r - -sr sin u« (2) 

If we assume that at t = 0, r = r , dr/dt = v , we find 

or in terms of hyperbolic functions, 

r = r cosh«t + ( — --^jsinhwt + ^-sinw* (-4) 

6.24. (a) Show that under suitable conditions the particle of Problem 6.23 can oscillate 
along the rod with simple harmonic motion and find these conditions, (b) What 
happens to the particle if the conditions of (a) are not satisfied? 

(a) The particle will oscillate with simple harmonic motion along the rod if and only if r = 
and v = 0/2<o. In this case, r = (g/2<* 2 ) sin at. Thus the amplitude and period of the simple 
harmonic motion in such case are given by g/2u> 2 and 2n-/w respectively. 

(6) If v - (g/2u) — ur then r = r- e _wt + (g/2a 2 ) smut and the motion is approximately simple 
harmonic after some time. Otherwise the mass will ultimately fly off the rod if it is finite. 

6.25. A projectile located at colatitude X is fired with velocity v in a southward direction 
at an angle « with the horizontal, {a) Find the position of the projectile after time t. 

(b) Prove that after time t the projectile is deflected toward the east of the original 
vertical plane of motion by the amount 

^<*g sin Xt z — w^o cos (a — X) t 2 

(a) We use the equations of Problem 6.15. Assuming the projectile starts at the origin, we have 

x = 0, y = 0, z = at * = U) 

Also, the initial velocity is v = v cos a i + v sin a k so that 

x = v cos a, y = 0, z = v sin « at t = (2) 



CHAP. 6] MOVING COORDINATE SYSTEMS 159 

Integrating equations (1), (2) and (3) of Problem 6.15, we obtain on using conditions (2), 

x = 2oj cos X y + v cos a (3) 

y — — 2(w cos X x + w sin X z) (4) 

z = —gt + 2w sin X y + v sin a (5) 

Instead of attempting to solve these equations directly we shall use the method of iteration 
or successive approximations as in Method 2 of Problem 6.16. Thus by integrating and using 
conditions (1), we find 

c 

x — 2wcosX I y du + (t; cosa)( (6) 



= — 2« cos X I x du — 2u sin X I z du 



(7) 



z = (v sin a)t — ±gt 2 + 2w sin X I y du (8) 

Jo 

As a first guess we use x = 0, y = 0, 2 = under the integral signs. Then (6), (7) and (8) 
become, neglecting terms involving w 2 , 

x = (v cos a)t (9) 

y = (10) 

z = (v sina)£ - \g& (11) 

To obtain a better guess we now use (9), (10) and (II) under the integral signs in (6), (7) and 
(8), thus arriving at 

X = (V COS a)t (12) 

y = —uv cos (a — \)t 2 + ^wgt 3 sin X (13) 

z — (i) sin a)t — Iflrt 2 (14) 

where we have again neglected terms involving w 2 . Further guesses again produce equations 
(12), (13) and (14), so that these equations are accurate up to terms involving « 2 . 

(6) From equation (13) we see that the projectile is deflected toward the east of the xz plane 
by the amount ^ugt s sin X — wi; cos (a — X) t 2 . If v = this agrees with Problem 6.16. 



6.26. Prove that when the projectile of Problem 6.25 returns to the horizontal, it will be at 

the distance 

o>vl sin 2 a 

— ;r-£ — (3 cos a cos A + sin a sin A) 

to the west of that point where it would have landed assuming no axial rotation 
of the earth. 

The projectile will return to the horizontal when z = 0, i.e., 

(v sin a)t — ^gt 2 — or t — (2v sin a)/g 

Using this value of t in equation (13) of Problem 6.25, we find the required result. 



160 MOVING COORDINATE SYSTEMS [CHAP. 6 

Supplementary Problems 

ROTATING COORDINATE SYSTEMS. VELOCITY AND ACCELERATION 

6.27. An xyz coordinate system moves with angular velocity «* = 2i — 3j + 5k relative to a fixed or 
inertial XYZ coordinate system having the same origin. If a vector relative to the xyz system 
is given as a function of time t by A = sin t i — cos t j + e -t k, find (a) dA/dt relative to 
the fixed system, (b) dA/dt relative to the moving system. 

Ans. (a) (6 cos t - 3e-*)i + (6 sin t - 2e-*)j + (3 sin t - 2 cos t - e-')k 
(b) cos t i + sin t j — e _t k 

6.28. Find d 2 A/dt 2 for the vector A of Problem 6.27 relative to (a) the fixed system and (6) the moving 

system. 

Ans. (a) (6 cos t - 45 sin t + Ue-^i + (40 cos * - 6 sin t - lle-^j 
+ (10 sin t - 23 cos t + 16e~')k 
(b) — sin t i + cos t j + e _t k 

6.29. An xyz coordinate system is rotating with angular velocity o» = 5i — 4j — 10k relative to a fixed 
XYZ coordinate system having the same origin. Find the velocity of a particle fixed in the xyz 
system at the point (3, 1, —2) as seen by an observer fixed in the XYZ system. 

Ans. 18i-20j + 17k 

6.30. Discuss the physical interpretation of replacing <* by -*> in (a) Problem 6.4, page 148, and 
(6) Problem 6.6, page 149. 

6.31. Explain from a physical point of view why you would expect the result of Problem 6.3, page 148, 
to be correct. 

6.32. An xyz coordinate system rotates with angular velocity « = cos t i + sin t j + k with respect to a 
fixed XYZ coordinate system having the same origin. If the position vector of a particle is given 
by r = sin t i - cos t j + tk, find (a) the apparent velocity and (6) the true velocity at any 
time t. Ans. (a) cos t i + sin t j + k (6) (t sin t + 2 cos *)i + (2 sin t - t cos t)j 

6.33. Determine (a) the apparent acceleration and (6) the true acceleration of the particle of 
Problem 6.32. 

Ans. (a) -sin ti + cos t j (6) (2t cost - 3 sin t)i + (3 cos t + 2t sin «)j + (1 - t)k 

CORIOLIS AND CENTRIPETAL ACCELERATIONS AND FORCES 

6.34. A ball is thrown horizontally in the northern hemisphere, (a) Would the path of the ball, 
if the Coriolis force is taken into account, be to the right or to the left of the path when it is not 
taken into account as viewed by the person throwing the ball? (6) What would be your answer 
to (a) if the ball were thrown in the southern hemisphere? Ans. (a) to the right, (6) to the left 

6.35. What would be your answer to Problem 6.34 if the ball were thrown at the north or south poles? 

6.36. Explain why water running out of a vertical drain will swirl counterclockwise in the northern 
hemisphere and clockwise in the southern hemisphere. What happens at the equator? 

6.37. Prove that the centrifugal force acting on a particle of mass m on the earth's surface is a 
vector (a) directed away from the earth and perpendicular to the angular velocity vector o> and 
(6) of magnitude mu> 2 R sin X where X is the colatitude. 

6.38. In Problem 6.37, where would the centrifugal force be (a) a maximum, (6) a minimum? 
Ans. (a) at the equator, (6) at the north and south poles. 

6.39. Find the centrifugal force acting on a train of mass 100,000 kg at (a) the equator (b) colatitude 30°. 
Ans. (a) 35.0 kg wt, (6) 17.5 kg wt 

6.40. (a) A river of width D flows northward with a speed v at colatitude X. Prove that the left bank 

of the river will be higher than the right bank by an amount equal to 

(2D u v cos\)(g 2 + 4« V cos^)- 1 ' 2 
where w is the angular speed of the earth about its axis. 
(6) Prove that the result in part (a) is for all practical purposes equal to (2Duv cos \)/g. 

6.41. If the river of Problem 6.40 is 2 km wide and flows at a speed of 5 km/hr at colatitude 45°, 
how much higher will the left bank be than the right bank? Ans. 2.9 cm 



CHAP. 6] MOVING COORDINATE SYSTEMS 161 

6.42. An automobile rounds a curve whose radius of curvature is p. If the coefficient of friction is /j, 
prove that the greatest speed with which it can travel so as not to slip on the road is Vwff- 

6.43. Determine whether the automobile of Problem 6.42 will slip if the speed is 60 mi/hr, n = .05 and 
(a) p = 500 ft, (6) p = 50 ft. Discuss the results physically. 

MOTION OF A PARTICLE RELATIVE TO THE EARTH 

6.44. An object is dropped at the equator from a height of 400 meters. If air resistance is neglected, 
how far will the point where it hits the earth's surface be from the point vertically below the 
initial position? Ans. 17.6 cm toward the east 

6.45. Work Problem 6.44 if the object is dropped (a) at colatitude 60° and (6) at the north pole. 
Ans. (a) 15.2 cm toward the east 

6.46. An object is thrown vertically upward at colatitude X with speed v . Prove that when it 
returns it will be at a distance westward from its starting point equal to (±av\ sin X)/Sg 2 . 

6.47. An object at the equator is thrown vertically upward with a speed of 60 mi/hr. How far from 
its initial position will it land? Ans. .78 inches 

6.48. With what speed must the object of Problem 6.47 be thrown in order that it return to a point 
on the earth which is 20 ft from its original position? Ans. 406 mi/hr 

6.49. An object is thrown downward with initial speed v . Prove that after time t the object is 
deflected east of the vertical by the amount 

wv sin X t 2 + ^ug sin X t 3 

6.50. Prove that if the object of Problem 6.49 is thrown downward from height h above the earth's 
surface, then it will hit the earth at a point east of the vertical at a distance 

^JP^ (Vv 2 + 2gh - v )H^v 2 Q + 2gh + 2v ) 

6.51. Suppose that the mass m of a conical pendulum of length I moves in a horizontal circle of 
radius a. Prove that (a) the speed is ayfg/^/l 2 — a 2 and (6) the tension in the string is 
mgly/l 2 — a 2 . 

6.52. If an object is dropped to the earth's surface prove that its path is a semicubical parabola. 

THE FOUCAULT PENDULUM 

6.53. Explain physically why the plane of oscillation of a Foucault pendulum should rotate clockwise 
when viewed from above the earth's surface in the northern hemisphere but counterclockwise in 
the southern hemisphere. 

6.54. How long would it take the plane of oscillation of a Foucault pendulum to make one complete 
revolution if the pendulum is located at (a) the north pole, (6) colatitude 45°, (c) colatitude 85°? 
Ans. (a) 23.94 hr, (6) 33.86 hr, (c) 92.50 hr 

6.55. Explain physically why a Foucault pendulum situated at the equator would not detect the 
rotation of the earth about its axis. Is this physical result supported mathematically? Explain. 

MOVING COORDINATE SYSTEMS IN GENERAL 

6.56. An xyz coordinate system rotates about the z axis with angular velocity o> = cos t i + sin t j 
relative to a fixed XYZ coordinate system where t is the time. The origin of the xyz system 
has position vector R = ti — j + t 2 k with respect to the XYZ system. If the position vector of 
a particle is given by r = (Zt + l)i — 2tj + 5k relative to the moving system, find the (a) apparent 
velocity and (6) true velocity at any time. 

6.57. Determine (a) the apparent acceleration and (6) the true acceleration of the particle in 
Problem 6.56. 

6.58. Work (a) Problem 6.5, page 148, and (b) Problem 6.7, page 149, if the position vector of the 
xyz system relative to the origin of the fixed XYZ system is R = t 2 i — 2tj + 5k. 



162 



MOVING COORDINATE SYSTEMS 



[CHAP. 6 



MISCELLANEOUS PROBLEMS 

6.59. Prove that due to the rotation o f the earth about its axis the app arent weight of an object of 
mass m at colatitude X is m^(g - u 2 R sin 2 x) 2 + ( w 2# s j n x cos X )2 w here R is the* radius of 
the earth. 



6.60. 



6.61. 



Prove that the angle /? which the apparent vertical at colatitude X makes with the true vertical 
o 2 R sin X cos X 



is given by tan /? = 



g — u 2 R sin 2 X ' 



Explain physically why the true vertical and apparent vertical would coincide at the equator 
and also the north and south poles. 



6.62. A stone is twirled in a vertical circle by a string of length 10 ft. Prove that it must have a 
speed of at least 20 ft/sec at the bottom of its path in order to complete the circle. 

6.63. A car C [Fig. 6-8] is to go completely around the vertical 
circular loop of radius a without leaving the track. 
Assuming the track is frictionless, determine the height 
H at which it must start. 

6.64. A particle of mass m is constrained to move on a friction- 
less vertical circle of radius a which rotates about a fixed 
diameter with constant angular speed w. Prove that the 
particle will make small oscillations about its equilibrium 
position with a frequency given by 2vau/-\/ a 2 w 4 — g 2 . 




6.65. Discuss what happens in Problem 6.64 if « = yfg/a. 

6.66. A hollow cylindrical tube AOB of length 2a [Fig. 6-9] 
rotates with constant angular speed « about a vertical 
axis through the center O. A particle is initially at rest 
in the tube at a distance & from O. Assuming no frac- 
tional forces, find (a) the position and (6) the speed of 
the particle at any time. 

6.67. (a) How long will it take the particle of Problem 6.66 to 
come out of the tube and (&) what will be its speed as 

it leaves? Ans. (a) — In (a + y/a 2 — b 2 ) 



Fig. 6-8 



CJ> 



~B 



Fig. 6-9 



6.68. Find the force on the particle of Problem 6.66 at any position in the tube. 

6.69. A mass, attached to a string which is suspended from a fixed point, moves in a horizontal circle 
having center vertically below the fixed point with a speed of 20 revolutions per minute. 
Find the distance of the center of the circle below the fixed point. Ans. 2.23 meters 

6.70. A particle on a frictionless horizontal plane at colatitude X is given an initial speed v in a 
northward direction. Prove that it describes a circle of radius v /(2w cos X) with period n7(<o cos X). 

6.71. The pendulum bob of a conical pendulum describes a horizontal circle of radius a. If the length 
of the pendulum is I, prove that the period is given by 4ir 2 y/l 2 — a 2 /g. 

6.72. A particle constrained to move on a circular wire of radius a and coefficient n is given an initial 
velocity v . Assuming no other forces act, how long will it take for the particle to come to rest? 



6.73. (a) Prove that if the earth were to rotate at an angular speed given by y2g/R where R is its 
radius and g the acceleration due to gravity, then the weight of a particle of mass m would be 
the same at all latitudes. (6) What is the numerical value of this angular speed? 
Ans. (b) 1.74 X 10~3 rad/sec 



CHAP. 6] 



MOVING COORDINATE SYSTEMS 



163 



6.74. A cylindrical tank containing water rotates about its axis with constant angular speed &> 
so that no water spills out. Prove that the shape of the water surface is a paraboloid of revolution. 

6.75. Work (a) Problem 6.16 and (6) Problem 6.17, accurate to terms involving w 2 . 

6.76. Prove that due to the earth's rotation about its axis, winds in the northern hemisphere traveling 
from a high pressure area to a low pressure area are rotated in a counterclockwise sense when 
viewed above the earth's surface. What happens to winds in the southern hemisphere? 



6.77. (a) Prove that in the northern hemisphere winds from the north, 
east, south and west are deflected respectively toward the west, 
north, east and south as indicated in Fig. 6-10. (b) Use this to 
explain the origin of cyclones. 

6.78. Find the condition on the angular speed so that a particle will 
describe a horizontal circle inside of a frictionless vertical cone 
of angle a. 

6.79. Work Problem 6.78 for a hemisphere. 

6.80. The period of a simple pendulum is given by P. Prove that its period when it is suspended from 
the ceiling of a train moving with speed v around a circular track of radius p is given by 

PVp~9/y/v* + p 2g2. 




6.81. Work Problem 6.25 accurate to terms involving w 2 . 

6.82. A thin hollow cylindrical tube OA inclined at angle a with the 
horizontal rotates about the vertical with constant angular speed 
w [see Fig. 6-11]. If a particle constrained to move in this tube 
is initially at rest at a distance a from the intersection O of the 
tube and the vertical axis of rotation, prove that its distance r 
from O at any time t is r = a cosh (ut cos a). 

6.83. Work Problem 6.82 if the rod has coefficient of friction /i. 

6.84. Prove that the particle of Problem 6.82 is in stable equilibrium 
between the distances from O given by 

g sin a ( 1 + n tan a 
w 2 \ tan a — fi 

assuming tan a < l//x. 



g sin a / l — jit tan a 
w 2 V tan a + n 



and 




Fig. 6-11 



6.85. 



6.86. 



A train having a maximum speed equal to v is to round a curve with radius of curvature p. Prove 
that if there is to be no lateral thrust on the outer track, then this track should be at a 
height above the inner track given by av\l yjv\ + P 2 g 2 where a is the distance between tracks. 

A projectile is fired at colatitude X with velocity v directed toward the west and at angle a with 

the horizontal. Prove that if terms involving « 2 are neglected, then the time taken to reach 

the maximum height is 

v sin a 2uv sin X sin a cos a 

g ff 2 

Compare with the case where w = 0, i.e. that the earth does not rotate about its axis. 



6.87. In Problem 6.86, prove that the maximum height reached is 

v 2 sin 2 a 2uv% sin X sin 2 a cos a 
2g g* 

Compare with the case where <o = 0. 



164 MOVING COORDINATE SYSTEMS [CHAP. 6 

6.88. Prove that the range of the projectile of Problem 6.86 is 

Vq sin 2a uVq sin a sin X (8 sin 2 a — 6) 



+ 



3^2 



Thus show that if terms involving co 2 and higher are neglected, the range will be larger, smaller 
or the same as the case where <o = 0, according as a > 60°, a < 60° or a — 60° respectively. 

6.89. If a projectile is fired with initial velocity v x i + v 2 j + v 3 k from the origin of a coordinate system 
fixed relative to the earth's surface at colatitude X, prove that its position at any later time t 
will be given by 

x — v x t + uv 2 t 2 cos X 

y — v 2 t — at 2 ^! cos \ + v 3 sin X) + ^ugt 3 sin X 
z — v 3 t — ±gt 2 + uv 2 t 2 sin X 
neglecting terms involving co 2 . 

6.90. Work Problem 6.89 so as to include terms involving co 2 but exclude terms involving co 3 . 

6.91. An object of mass m initially at rest is dropped from height h to the earth's surface at colatitude X. 
Assuming that air resistance proportional to the instantaneous speed of the object is taken into 
account as well as the rotation of the earth about its axis, prove that after time t the object is 
deflected east of the vertical by the amount 

2t ° ^ 3 " X [(g ~ 2hp 2 )(l - e-Pt) + phte-Bt - pgt + \g$ 2 t 2 \ 
neglecting terms of order w 2 and higher. 

6.92. Work Problem 6.91, obtaining accuracy up to and including terms of order u 2 . 

6.93. A frictionless inclined plane of length I and angle a located at colatitude X is so situated that 
a particle placed on it would slide under the influence of gravity from north to south. If the 
particle starts from rest at the top, prove that it will reach the bottom in a time given by 



y g sin a 



2ul sin X cos a 



3g 
and that its speed at the bottom is 

yj2gl sin a — f ul sin a cos a sin X 
neglecting terms of order co 2 . 

6.94. (a) Prove that by the time the particle of Problem 6.93 reaches the bottom it will have undergone 
a deflection of magnitude 



2Zco / 21 , ... 

-5- \ : COS (a + X) 

o \ g sin a 



to the east or west respectively according as cos (a + X) is greater than or less than zero. 
(&) Discuss the case where cos (a + X) = 0. (c) Use the result of (a) to arrive at the result 
of Problem 6.17. 

6.95. Work Problems 6.93 and 6.94 if the inclined plane has coefficient of friction (i. 



Chapter 7 SYSTEMS 

of PARTICLES 



DISCRETE AND CONTINUOUS SYSTEMS 

Up to now we have dealt mainly with the motion of an object which could be considered 
as a particle or point mass. In many practical cases the objects with which we are concerned 
can more realistically be considered as collections or systems of particles. Such systems 
are called discrete or continuous according as the particles can be considered as separated 
from each other or not. 

For many practical purposes a discrete system having a very large but finite number 
of particles can be considered as a continuous system. Conversely a continuous system can 
be considered as a discrete system consisting of a large but finite number of particles. 

DENSITY 

For continuous systems of particles occupying a region of space it is often convenient to 
define a mass per unit volume which is called the volume density or briefly density. 
Mathematically, if AM is the total mass of a volume At of particles, then the density can 

be defined as ,. aM m 

<t = hm — — v-U 

at^o At 

The density is a function of position and can vary from point to point. When the density 
is a constant, the system is said to be of uniform density or simply uniform. 

When the continuous system of particles occupy a surface, we can similarly define a 
surface density or mass per unit area. Similarly when the particles occupy a line [or curve] 
we can define a mass per unit length or linear density. 

RIGID AND ELASTIC BODIES 

In practice, forces applied to systems of particles will change the distances between 
individual particles. Such systems are often called deformable or elastic bodies. In some 
cases, however, deformations may be so slight that they may for most practical purposes 
be considered non-existent. It is thus convenient to define a mathematical model in which 
the distance between any two specified particles of a system remains the same regardless 
of applied forces. Such a system is called a rigid body. The mechanics of rigid bodies is 
considered in Chapters 9 and 10. 

DEGREES OF FREEDOM 

The number of coordinates required to specify the position of a system of one or more 
particles is called the number of degrees of freedom of the system. 

Example 1. 

A particle moving freely in space requires 3 coordinates, e.g. (x, y, z), to specify its position. Thus 
the number of degrees of freedom is 3. 

165 



166 



SYSTEMS OF PARTICLES 



[CHAP. 7 



Example 2. 

A system consisting of N particles moving freely in space requires 3N coordinates to specify its 
position. Thus the number of degrees of freedom is 32V. 

A rigid body which can move freely in space has 6 degrees of freedom, i.e. 6 coordinates 
are required to specify the position. See Problem 7.2. 



CENTER OP MASS 

Let n, r 2 , . . . , r N be the position vectors of a system of N particles of masses 

m u m 2 , . . .,m N respectively [see Fig. 7-1]. 

The center of mass or centroid of the system of particles is defined as that point C 
having position vector 

mxxx + m 2 r 2 + • • • + m N r N _ 

~ M 



r = 



1 N 



(2) 



m 1 + m 2 + • • • + m N 

N 

where M = ^ m v is the total mass of the system. We sometimes use 2 or simply J) 
in place of 2 • 

..«2 





Fig. 7-1 



Fig. 7-2 



For continuous systems of particles occupying a region % of space in which the 
volume density is o-, the center of mass can be written 






■rdi 



r = 



J or dr 



where the integral is taken over the entire region % [see Fig. 7-2]. If we write 

f = xi + yj + zk, r„ = x v i + y v j + z v k 
then (3) can equivalently be written as 



and 



- _ S WyXy _ 2) m v y v 

X ~ M > $ ~ M ' 

JaXdr I aydr 

<V - *J CD 



X = 



% 



„-, _ *"K 



M ' ° M 

where the total mass is given by either 

M = ^m v 

= \ <rdr 



z = 



M 

JaZdr 



M 



M 



(5) 

(6) 
(7) 



CHAP. 7] SYSTEMS OF PARTICLES 167 

The integrals in (3), (5) or (7) can be single, double or triple integrals, depending on 
which may be preferable. 

In practice it is fairly simple to go from discrete to continuous systems by merely 
replacing summations by integrations. Consequently we will present all theorems for 
discrete systems. 



CENTER OF GRAVITY 

If a system of particles is in a uniform gravitational field, the center of mass is some- 
times called the center of gravity. 

MOMENTUM OF A SYSTEM OF PARTICLES 

If v, = dxjdt = r„ is the velocity of m v , the total momentum of the system is defined as 

N N 

p = ]£ ra„v„ = ^m v r v (8) 

v=l v=l 

We can show [see Problem 7.3] that 

p = Mv = M^ = Mi (9) 

where v = di/dt is the velocity of the center of mass. 

This is expressed in the following 

Theorem 7.1. The total momentum of a system of particles can be found by multiplying 
the total mass M of the system by the velocity v of the center of mass. 



MOTION OF THE CENTER OF MASS 

Suppose that the internal forces between any two particles of the system obey Newton's 
third law. Then if F is the resultant external force acting on the system, we have [see 
Problem 7.4] , „ ,_ 

F = ft = M w> = M dt < I0) 

This is expressed in 

Theorem 7.2. The center of mass of a system of particles moves as if the total mass 
and resultant external force were applied at this point. 



CONSERVATION OF MOMENTUM 

Putting F = in (10), we find that 

N 

p = ^ m v \ v = constant (11) 

v=l 

Thus we have 

Theorem 7.3. If the resultant external force acting on a system of particles is zero, 
then the total momentum remains constant, i.e. is conserved. In such case the center of mass 
is either at rest or in motion with constant velocity. 

This theorem is often called the principle of conservation of momentum. It is a generaliza- 
tion of Theorem 2-8, page 37. 



168 SYSTEMS OF PARTICLES [CHAP. 7 

ANGULAR MOMENTUM OF A SYSTEM OF PARTICLES 

The quantity N 

O = 2 m v {r„Xv v ) (12) 

is called the total angular momentum [or moment of momentum] of the system of particles 
about origin 0. 

THE TOTAL EXTERNAL TORQUE ACTING ON A SYSTEM 

If F„ is the external force acting on particle v, then r„ x F„ is called the moment of the 
force F v or torque about 0. The sum 

N 

A = 2 r v X F„ (iS) 

is called the £o£aZ external torque about the origin. 

RELATION BETWEEN ANGULAR MOMENTUM AND 
TOTAL EXTERNAL TORQUE 

If we assume that the internal forces between any two particles are always directed 
along the line joining the particles [i.e. they are central forces], then we can show as in 
Problem 7.12 that 

da 

A = dt W 

Thus we have 

Theorem 7.4. The total external torque on a system of particles is equal to the time 
rate of change of the angular momentum of the system, provided the internal forces 
between particles are central forces. 

CONSERVATION OF ANGULAR MOMENTUM 

Putting A = in (U), we find that 

N 

O = ^ m v (r v x v„) = constant (15) 

Thus we have 

Theorem 7.5. If the resultant external torque acting on a system of particles is zero, 
then the total angular momentum remains constant, i.e. is conserved. 

This theorem is often called the principle of conservation of angular momentum. It is the 
generalization of Theorem 2.9, page 37. 

KINETIC ENERGY OF A SYSTEM OF PARTICLES 

The total kinetic energy of a system of particles is defined as 

\ N j N 

T = « Jm,t)J = sXm,^ (16) 

WORK 

If Tv is the force (external and internal) acting on particle v, then the total work done 
in moving the system of particles from one state [symbolized by 1] to another [symbolized 

by 2] is 

W12 = £ I yWr„ (17) 

v=l *s \ 



CHAP. 7] SYSTEMS OF PARTICLES 169 

As in the case of a single particle, we can prove the following 

Theorem 7.6. The total work done in moving a system of particles from one state 
where the kinetic energy is 2\ to another where the kinetic energy is T 2 , is 

W12 = T 2 -T, (is) 

POTENTIAL ENERGY. CONSERVATION OF ENERGY 

When all forces, external and internal, are conservative, we can define a total potential 
energy V of the system. In such case we can prove the following 

Theorem 7.7. If T and V are respectively the total kinetic energy and total potential 
energy of a system of particles, then 

T + V = constant (19) 

This is the principle of conservation of energy for systems of particles. 

MOTION RELATIVE TO THE CENTER OF MASS 

It is often useful to describe the motion of a system of particles about [or relative to] 
the center of mass. The following theorems are of fundamental importance. In all cases 
primes denote quantities relative to the center of mass. 

Theorem 7.8. The total linear momentum of a system of particles about the center 
of mass is zero. In symbols, 

N N 

Y, m v< = ^m v r f v = (20) 

v=X v=l 

Theorem 7.9. The total angular momentum of a system of particles about any point 
equals the angular momentum of the total mass assumed to be located at the center of mass 
plus the angular momentum about the center of mass. In symbols, 

O = r X Mv + 2 m v (r' v X v' v ) (21) 

v=l 

Theorem 7.10. The total kinetic energy of a system of particles about any point O 
equals the kinetic energy of translation of the center of mass [assuming the total mass 
located there] plus the kinetic energy of motion about the center of mass. In symbols, 

T = ^Mv 2 + |5>X 2 (22) 

Theorem 7.11. The total external torque about the center of mass equals the time rate 
of change in angular momentum about the center of mass, i.e. equation (lb) holds not 
only for inertial coordinate systems but also for coordinate systems moving with the 
center of mass. In symbols, 

da' 

A = -W M 

If motion is described relative to points other than the center of mass, the results 
in the above theorems become more complicated. 



IMPULSE 

If F is the total external force acting on a system of particles, then 



170 SYSTEMS OF PARTICLES [CHAP. 7 



f 

•St. 



¥dt (U) 

h 

is called the total linear impulse or briefly total impulse. As in the case of one particle, 
we can prove 

Theorem 7.12. The total linear impulse is equal to the change in linear momentum. 

Similarly if A is the total external torque applied to a system of particles about origin O, 
then t2 

( A eft (25) 

is called the total angular impulse. We can then prove 

Theorem 7.13. The total angular impulse is equal to the change in angular momentum. 



CONSTRAINTS. HOLONOMIC AND NON-HOLONOMIC CONSTRAINTS 

Often in practice the motion of a particle or system of particles is restricted in some 
way. For example, in rigid bodies [considered in Chapters 9 and 10] the motion must be such 
that the distance between any two particular particles of the rigid body is always the same. 
As another example, the motion of particles may be restricted to curves or surfaces. 

The limitations on the motion are often called constraints. If the constraint condition 
can be expressed as an equation 

<£(ri,r 2 , ...,r w , t) = (26) 

connecting the position vectors of the particles and the time, then the constraint is called 
holonomic. If it cannot be so expressed it is called non-holonomic. 



VIRTUAL DISPLACEMENTS 

Consider two possible configurations of a system of particles at a particular instant 
which are consistent with the forces and constraints. To go from one configuration to 
the other, we need only give the vth particle a displacement Sr* from the old to the new 
position. We call Sr, a virtual displacement to distinguish it from a true displacement 
[denoted by dr v ] which occurs in a time interval where forces and constraints could be 
changing. The symbol 8 has the usual properties of the differential d; for example, 
S(sin0) = cos0 80. 

STATICS OF A SYSTEM OF PARTICLES. 
PRINCIPLE OF VIRTUAL WORK 

In order for a system of particles to be in equilibrium, the resultant force acting on each 
particle must be zero, i.e. F„ = 0. It thus follows that F, • 8r v = where F v • 8r v is called 
the virtual work. By adding these we then have 

2Fv8r„ = (27) 

v = l 

If constraints are present, then we can write 

F„ = F l v a) + F< c) (28) 

where F< a) and F< c) are respectively the actual force and constraint force acting on the vth 
particle. By assuming that the virtual work of the constraint forces is zero [which is true 
for rigid bodies and for motion on curves and surfaces without friction], we arrive at 



CHAP. 7] SYSTEMS OF PARTICLES 171 

Theorem 7.14. A system of particles is in equilibrium if and only if the total virtual 
work of the actual forces is zero, i.e. if 

2 F< 0) • 8r„ = (29) 

v=l 

This is often called the principle of virtual work. 



EQUILIBRIUM IN CONSERVATIVE FIELDS. 
STABILITY OF EQUILIBRIUM 

The results for equilibrium of a particle in a conservative force field [see page 38] 
can be generalized to systems of particles. The following theorems summarize the basic 
results. 

Theorem 7.15. If V is the total potential of a system of particles depending on 
coordinates q v q 2 , . . . , then the system will be in equilibrium if 

W^ ' 8o- 2 = °> •• («) 

Since the virtual work done on the system is 

" = £". + £".♦••• 

(31) is equivalent to the principle of virtual work. 

Theorem 7.16. A system of particles will be in stable equilibrium if the potential V 
is a minimum. 



In case V depends on only one coordinate, say q v sufficient conditions are 

dV A dW 

Other cases of equilibrium where the potential is not a minimum are called unstable. 



D'ALEMBERT'S PRINCIPLE 

Although Theorem 7.14 as stated applies to the statics of a system of particles, it can be 
restated so as to give an analogous theorem for dynamics. To do this we note that according 
to Newton's second law of motion, 

F„ = p„ or F„ - p„ = (so) 

where p v is the momentum of the vth particle. The second equation amounts to saying 
that a moving system of particles can be considered to be in equilibrium under a force 
F„ - p v , i.e. the actual force together with the added force -p„ which is often called the 
reversed effective force on particle v. By using the principle of virtual work we can then 
arrive at 

Theorem 7.17. A system of particles moves in such a way that the total virtual work 

2 (F^ a) - p.) • 8r„ = 



With this theorem, which is often called D'Alembert's principle, we can consider dynamics 
as a special case of statics. 



172 SYSTEMS OF PARTICLES [CHAP. 7 



Solved Problems 

DEGREES OF FREEDOM 

7.1. Determine the number of degrees of freedom in each of the following cases: 
(a) a particle moving on a given space curve; (b) five particles moving freely in a 
plane; (c) five particles moving freely in space; (d) two particles connected by a 
rigid rod moving freely in a plane. 

(a) The curve can be described by the parametric equations x = x(s), y = y(s), z = «(s) where 
s is the parameter. Then the position of a particle on the curve is determined by specifying 
one coordinate, and hence there is one degree of freedom. 

(b) Each particle requires two coordinates to specify its position in the plane. Thus 5 • 2 = 10 
coordinates are needed to specify the positions of all 5 particles, i.e. the system has 10 degrees 
of freedom. 

(c) Since each particle requires three coordinates to specify its position, the system has 5*3 = 15 
degrees of freedom. 

(d) Method 1. 

The coordinates of the two particles can be expressed by {x lf y{) and (x 2 ,y 2 ), i.e. a total 
of 4 coordinates. However, since the distance between these points is a constant a [the length 
of the rigid rod], we have (x 1 — x 2 ) 2 + (l/i — 2/2) 2 = ft2 so that one of the coordinates can be 
expressed in terms of the others. Thus there are 4 — 1 = 3 degrees of freedom. 

Method 2. 

The motion is completely specified if we give the two coordinates of the center of mass 
and the angle made by the rod with some specified direction. Thus there are 2 + 1 = 3 degrees 
of freedom. 

72. Find the number of degrees of freedom for a rigid body which (a) can move freely 
in three dimensional space, (b) has one point fixed but can move in space about 
this point. 

(a) Method 1. 

If 3 non-collinear points of a rigid body are fixed in space, then the rigid body is also fixed 
in space. Let these points have coordinates (x lt y x , zj, {x 2 , y 2 , z 2 ), (x 3 , y 3 , z 3 ) respectively, a total 
of 9. Since the body is rigid we must have the relations 
(x x - x 2 ) 2 + (y x - y 2 ) 2 + (z t - z 2 ) 2 = constant, (x 2 - x 5 ) 2 + (y 2 - y 3 ) 2 + {z 2 - z 3 ) 2 = constant, 

(x 3 - x x ) 2 + (y 3 - y x ) 2 + (z 3 - z t ) 2 = constant 
hence 3 coordinates can be expressed in terms of the remaining 6. Thus 6 independent 
coordinates are needed to describe the motion, i.e. there are 6 degrees of freedom. 

Method 2. 

To fix one point of the rigid body requires 3 coordinates. An axis through this point 
is fixed if we specify 2 ratios of the direction cosines of this axis. A rotation about the axis 
can then be described by 1 angular coordinate. The total number of coordinates required, 
i.e. the number of degrees of freedom, is 3 + 2 + 1 = 6. 

(6) The motion is completely specified if we know the coordinates of two points, say (x^y^Zj) 
and (x 2 ,y 2 ,z 2 ), where the fixed point is taken at the origin of a coordinate system. But since 
the body is rigid we must have 

x\ + y\ + z\ = constant, x\ + yl + z\= constant, (x x - x 2 ) 2 + (y t - y 2 ) 2 + {z t - z 2 ) 2 = constant 
from which 3 coordinates can be found in terms of the remaining 3. Thus there are 3 degrees 
of freedom. 

CENTER OF MASS AND MOMENTUM OF A SYSTEM OF PARTICLES 

7.3. Prove Theorem 7.1, page 167: The total momentum of a system of particles can be 

found by multiplying the total mass M of the system by the velocity v of the center 

of mass. 



CHAP. 7] SYSTEMS OF PARTICLES 173 

2 rn v v v 
The center of mass is by definition, f = ^ — . 

Then the total momentum is p — 2 i^ v v„ — 2 »»,, r„ = M dr/dt = Mx. 

7.4. Prove Theorem 7.2, page 167: The center of mass of a system of particles moves 
as if the total mass and resultant external force were applied at this point. 

Let F„ be the resultant external force acting on particle v while f„ x is the internal force on 
particle v due to particle \. We shall assume that f„„ = 0, i.e. particle v does not exert any 
force on itself. 

By Newton's second law the total force on particle v is 

„ dPy J2 

F " + ? f * = ~dt = dP> (w » r "> W 

where the second term on the left represents the resultant internal force on particle v due to all 
other particles. 



Summing over v in equation (1), we find 

if. — 

d& 



2 F v + 2 2 t v \ = m i 2 m v r v I (2) 



Now according to Newton's third law of action and reaction, f„ x = — f Xl> so that the double 
summation on the left of {2) is zero. If we then write 

F = 2 F„ and f = ^^ m v r v (3) 

V 1Y1 V 

d 2 r 
(2) becomes F = M-r^ (4) 

Since F is the total external force on all particles applied at the center of mass f, the required 
result is proved. 

7.5. A system of particles consists of a 3 gram mass located at (1, 0, -1), a 5 gram mass 
at (-2, 1, 3) and a 2 gram mass at (3, -1, 1). Find the coordinates of the center of mass. 

The position vectors of the particles are given respectively by 

r t = i - k, r 2 = -2i + j + 3k, r 3 = 3i - j + k 
Then the center of mass is given by 



f _ 3(i - k) + 5(-2i + j + 3k) + 2(3i - j + k) _ 1 . , 3 . , 7 

3 + 5 + 2 

Thus the coordinates of the center of mass are (—J*. A. 1) 

v 1 0' 1 o ' 



3 + 5 + 2 10 1 + 10 j + 5 k 



7.6. Prove that if the total momentum of a system is constant, i.e. is conserved, then the 
center of mass is either at rest or in motion with constant velocity. 
The total momentum of the system is given by 

d ^ .. d r2^r„] rff 



Then if p is constant, so also is dr/dt, the velocity of the center of mass. 

7.7. Explain why the ejection of gases at high velocity from the rear of a rocket will move 
the rocket forward. 

Since the gas particles move backward with high velocity and since the center of mass does 
not move, the rocket must move forward. For applications involving rocket motion, see Chapter 8. 



174 



SYSTEMS OF PARTICLES 



[CHAP. 7 



Ar„ = Ax v Ay v &z v 



7.8. Find the centroid of a solid region % as in Fig. 7-3. 

Consider the volume element At v of the solid. The 
mass of this volume element is 

AM V = o v At v = a v Ax v Ay v Az v 

where a v is the density [mass per unit volume] and 
Ax v , Ay v , Az„ are the dimensions of the volume element. 
Then the centroid is given approximately by 

2 r„ AM V 2 r v a v Ar v 2 *„ a v Ax v Ay v Az v 

2 AM V 2 <r v At v 2 °v ^ x v ^Vv ^ z v 

where the summation is taken over all volume elements 
of the solid. 

Taking the limit as the number of volume elements becomes infinite in such a way that 
Ar„ -» or Ax v -» 0, Ay v -» 0, Az v -» 0, we obtain for the centroid of the solid: 

dM | radr JJJ radxdydz 

_ % 




Fig. 7-3 









9. 



dx dy dz 



where the integration is to be performed Over %, as indicated. 

Writing r = xi + yj + zk, f = xi + yj + zk, this can also be written in component form as 

\\\ yadxdydz jJj zadxdydz 

% = _ 



III x a dx dy dz 

% 



% 



III adxdy dz 
9 



Jlf* 



dx dy dz 



sss- 



dx dy dz 



7.9. Find the centroid of the region bounded by the plane x + y + z = a and the planes 
x = 0, y = 0, z = 0. 

The region, which is a tetrahedron, is indicated in Fig. 7-4. To find the centroid, we use 
the results of Problem 7.8. 

In forming the sum over all volume elements of the region, it is advisable to proceed in an 
orderly fashion. One possibility is to add first all terms corresponding to volume elements 
contained in a column such as PQ in the figure. This amounts to keeping x v and y v fixed and 
adding over all z v . Next keep *„ fixed but sum over all y v . This amounts to adding all columns, 
such as PQ, contained in a slab RS, and consequently amounts to summing over all cubes contained 
in such a slab. Finally, vary x v . This amounts to addition of all slabs such as RS. 

In performing the integration over %, we use these 
same ideas. Thus keeping x and y constant, integrate 
from z = [base of column PQ] to z — a — x — y [top 
of column PQ]. Next keep x constant and integrate 
with respect to y. This amounts to addition of columns 
having bases in the xy plane [z = 0] located anywhere 
from R [where y = 0] to S [where x + y = a or y = 
a — x], and the integration is from y = to y = a — x. 
Finally, we add all slabs parallel to the yz plane, which 
amounts to integration from x — to x — a. We 
thus obtain 



At„ = Ax v Ay v Az v 



t = 



Jr>a pa — x pa—x — y 
I I a(xi + yj + zk) dz dy dx 

x = *A/ = *z = 



a — x /*a — x—i 



na pa — x s* 
K 'x=Q *^j/ = *^=0 



a dz dy dx 




Fig. 7-4 



Since a is constant in this case, it may be cancelled. The denominator without a is evaluated 
to be a 3 /6, and the numerator without a is (a 4 /24)(i + j + k). Thus the center of mass is 
f = (a/4)(i + j + k) or x = a/4, y = a/4, z = a/4. 



CHAP. 7] 



SYSTEMS OF PARTICLES 



175 



7.10. Find the centroid of a semi-circular region of radius a. 

Method 1. Using rectangular coordinates. 

Choose the region as in Fig. 7-5. The equation of the circle C is x 2 + y 2 = a 2 or y = y/a 2 - x 2 
since y ^ 0. 

If a is the mass per unit area, assumed constant, then the coordinates of the centroid 
are given by 



3 = 



J xadA JJ xdydx j j 



y/l^x* 



xdy dx 



V = 






f <rdA ff dydx C C 



Jtf^ 



= 



dydx 



ydydx 



/. 



x = —a 3/=0 



ydydx 



odA 



ss 



dy dx 



J-._ J 



V^? 



dydx 



2a 3 /3 
ira 2 /2 



4a 

3lT 



a "y=0 



Note that we can write x - immediately, since by symmetry the centroid is on the y axis. 
The denominator for y can be evaluated without integrating by noting that it represents the 
semi-circular area which is %ira 2 . 





Fig. 7-5 



Fig. 7-6 



Method 2. Using polar coordinates. 

The equation of the circle is r = a [see Fig. 7-6]. As before, we see by symmetry that the 
centroid must lie on the y axis, so that x = 0. Since y = r sin e and dA=rdrde in polar 
coordinates, we can write 



y = 



yadA I \ (r sin 9 ) r dr de 



s 



dA 



% 



r=0 



rdrde 



2q3/3 



Aa 
Sir 



7.11. Find the center of mass of a uniform solid hemisphere of radius a. 

By symmetry the center of mass lies on the 
z axis [see Fig. 7-7]. Subdivide the hemisphere 
into solid circular plates of radius r, such as 
ABCDEA. If the center G of such a ring is at 
distance z from the center O of the hemisphere, 
r 2 + z 2 = a 2 . Then if dz is the thickness of the 
plate, the volume of each ring is 

irr 2 dz = ir{a 2 -z 2 )dz 

and the mass is ira(a 2 — z 2 ) dz. Thus we have 



£ 



traz{a 2 — z 2 ) dz 



2 — z = 



J a 
va(a 2 — z 2 ) dz 



= la 




z=0 



Fig. 7-7 



176 SYSTEMS OF PARTICLES [CHAP. 7 

ANGULAR MOMENTUM AND TORQUE 

7.12. Prove Theorem 7.4, page 168: The total external torque on a system of particles is 
equal to the time rate of change of angular momentum of the system, provided that 
the internal forces between particles are central forces. 
As in equation (1) of Problem 7.4, we have 



d P v d 

x 

Multiplying both sides of (1 ) by r v X , we have 



f„ + 2f„x = -of = Tt {m " Vv) {1) 



r,XF v + 2 r„ X f„ x = v v X ^(m„ v„) (*) 

d d 

Since r„ X ^(m„v„) = ^K(r,Xv v )} (3) 



(2) becomes r,XF, + 2 *„ X f vX = ^K(r,Xv,)} (4) 

Summing over j» in (4), we find 



2r,XF, + 2 2r„Xf vX = j- i 2 m v (t v X v„) f (5) 

l> l> X CtC ^ j; J 

Now the double sum in (5) is composed of terms such as 

r v XU + *xXf x „ («) 

which becomes on writing £ X|/ = -f„ x according to Newton's third law, 

r„Xf„ x - r x Xf„ x = (r„ - r x ) X f vX (7) 

Then since we suppose that the forces are central, i.e. f„ x has the same direction as r„ — r x , it follows 
that (7) is zero and also that the double sum in (5) is zero. Thus equation (5) becomes 

2r v XF„ = i\^m v (r v Xv v )\ or A = d ° 



dt \f "" v ^ v " *" ; J " dt 

where A = 2 *i/ X F„, O = 2 Wj/(r„ X v„). 



WORK, KINETIC ENERGY AND POTENTIAL ENERGY 

7.13. Prove Theorem 7.6, page 169: The total work done in moving a system of particles 
from one state to another with kinetic energies 2\ and T 2 respectively is T2-T1. 
The equation of motion of the pth particle in the system is 

(1) 



(2) 



T v 


= F„ + 2U = 


d / • X 

^(m,r,) 


Taking the dot product of both sides with r„, we have 




Tv'K == 


F„ • r v + 2 t v \ ' K z 


d i • x 


Since r„ • -^ (ra„ r„) 


Id, ,• • >-. 


Id/ 2\ 

= 2 5t M 


(2) can be written 

Tv'K = 


F„ • r„ + 2 f i>x * K 

X 


Id/ 2\ 
= 2di {m ^ 



(3) 



2^-r, = 2F„.r„ + 22U-r, = ^(2™,^) U) 



Summing over y in equation (5), we find 



CHAP. 7] SYSTEMS OF PARTICLES 177 

Integrating both sides of (4) with respect to t from t = t x to t = t 2 , we find 

w 12 = 2 ('r v 'r v dt = 2 ( 2 F v -l v dt + 22 ( 2 t vX -r v dt 

ti »i l l 

Using the fact that r„ <Zt = dr„ and the symbols 1 and 2 for the states at times t x and t 2 
respectively, this can be written 

W 12 = 2 Cr v 'dr v = 2 ( 2 F v 'dr v + 22 f f„x ' *„ = T 2 - T x (5) 

where Ti and T^ are the total kinetic energies at t x and t 2 respectively. Since 



W 



= 2 f f v 'dr v (6) 



12 

is the total work done (by external and internal forces) in moving the system from one state to 
another, the required result follows. 

It should be noted that the double sum in (5) indicating work done by the internal forces, 
cannot be reduced to zero either by using Newton's third law or the assumption of central forces. 
This is in contradistinction to the double sums in Problems 7.4 and 7.12 which can be reduced to zero. 



7.14. Suppose that the internal forces of a system of particles are conservative and are 
derived from a potential 

Vxv (rw) = V v \ (r v \) 

where n„ = r v \ = V( x * ~ x ») 2 + (#*. — Vv) 2 + (z\ - z v ) 2 is the distance between par- 
ticles A and v of the system. 

(a) Prove that J Jf^-dr, = -« 2 2 dVw where f x „ is the internal force on 

v X £t v \ 

particle v due to particle A. 

(b) Evaluate the double sum 22 f f\v'dr v of Problem 7.13. 

(a) The force acting on particle v is 

Wxv. Wxv. dV Xv 

The force acting on particle X is 

BV Xv ^ dV Xv dV Xv 

hv - ~~B^ 1 ~ ~d^ j ~ ~J^ k = ~S^d x V Xv = -V x V Xv = -i vX (2) 

The work done by these forces in producing the displacements dr v and dr x of particles v and X 
respectively is 



i vX 'dr v + i Xv 'dr x = 



JdV Xv d v Kv dV Xv dV Xv dV Xv dV Xv , -] 



= ~dV Xv 
Then the total work done by the internal forces is 

22fx„-rfr l , = -§22<n\, (s) 

v X * v \ 

the factor £ on the right being introduced because otherwise the terms in the summation 
would enter twice. 



178 



SYSTEMS OF PARTICLES 



[CHAP. 7 



(6) By integrating (3) of part (a), we have 

-2 



2 2) t Xy • dr v = - - 2 2 I 



C*Vx 



where y< lnt) and y 2 mt) denote the total internal potentials 



J22v x 



at times ^ and £ 2 respectively. 



2 t 



1/(int) -r/(int) 

K l K 2 



U) 



(5) 



7.15. Prove that if both the external and internal forces for a system of particles are 
conservative, then the principle of conservation of energy is valid. 

If the external forces are conservative, then we have 

F„ - -VV„ 



(1) 



from which 



2 (\-dr v - -2 fV, = v? 



T/(ext) 
K 2 



where V[ ext) and y< ext) denote the total external potential 

2V„ 

V 

at times t t and £ 2 respectively. 

Using {2) and equation (4) of Problem 7.14(6) in equation (5) of Problem 7.13, we find 

t 2 - t x = y{ ext) - v< ext) + v< Int) - y< int) = v x - y 2 (*) 



where 



Vi 



V 



(ext) 



+ vj 



(int) 



and 



Vo 



y^ ext) + y 2 



(int) 



U) 



are the respective total potential energies [external and internal] at times t x and t % . We thus find 
from (5), 

T t + V x - T 2 + y 2 or T + V = constant (5) 

which is the principle of conservation of energy. 



MOTION RELATIVE TO THE CENTER OF MASS 

7.16. Let r^ and \' v be respectively the position vector and velocity of particle v relative to 
the center of mass. Prove that (a) ^ m v r' v = 0, (b) ^ m v \' v = 0. 

V v 

(a) Let r^ be the position vector of particle v relative to O 
and f the position vector of the center of mass C relative 
to O. Then from the definition of the center of mass, 




r = ^^m v r v (1) 

where M — 2 m v From Fig. 7-8 we have 

r v = r ' v + f (2) Fig. 7-8 

Then substituting (2) into (1), we find 

from which 2 m v r v ~ ® 

v 

(b) Differentiating both sides of (3) with respect to t, we have 2 m* x v = °- 



(S) 



CHAP. 7] SYSTEMS OF PARTICLES 179 

7.17. Prove Theorem 7.9, page 169: The total angular momentum of a system of particles 
about any point O equals the angular momentum of the total mass assumed to be 
located at the center of mass plus the angular momentum about the center of mass. 

Let r„ be the position vector of particle v relative to O, f the position vector of the center of 
mass C relative to O and r' the position vector of particle v relative to C. Then 

*v = <+* tt) 

Differentiating with respect to t, we find 

v„ = r„ = i' v + r = \' v + v (2) 

where v is the velocity of the center of mass relative to O, v„ is the velocity of particle v relative to 
O, and v' v is the velocity of particle v relative to C. 

The total angular momentum of the system about is 

a = Sw.fr.Xv,,) = 2^{« + f)X( V ; + v)} 

V V 

= 2 m v (r' v Xvp + 2 m,(r' X v) + 2 m v (t X V v ) + 2 m v {l X v) (3) 

V V V V 

Now by Problem 7.16, 

K(<Xv) = J2^v<fXv = 



2 m v {f Xv) = f X -j 2 rn p \'V = 

2 m v (t X v) = < 2 «H, Mr X v) = M{i X v) 

Then (3) becomes, as required, 

O = 2m„(<Xv') + Af(fXv) 

7.18. Prove Theorem 7.10, page 169: The total kinetic energy of a system of particles 
about any point O equals the kinetic energy of the center of mass [assuming the 
total mass located there] plus the kinetic energy of motion about the center of mass. 
The kinetic energy relative to O [see Fig. 7-8] is 

* v *• V 

Using equation (2) of Problem 7.16 we find 

r v = r + r„ = v + v„ 
Thus (1) can be written 

T = ^2^{(v + v;).(v + V ;)} 

6 v 

= «2^v-v + 2™„v'< + h^^K'K 

*• v v & v v 

= |(?^)v- + v.jsm.v;} + §2*^; 

= gMv 2 + - 2 m„v' v 
since 2 ™> v v ' v ~ by Problem 7.16. 



,2 



180 SYSTEMS OF PARTICLES [CHAP. 7 

IMPULSE 

7.19. Prove Theorem 7.12: The total linear impulse is equal to the change in linear 
momentum. 

The total external force by equation (4) of Problem 7.4 is 

_, ifdH ,.dv 

F = M -di? = M lt 
Then the total linear impulse is 

X* 2 C H dv _ 

Ydt = I M ~^ dt ~ M *2 - M\ 1 = p 2 - Pi 

where p x = Mx t and p 2 — Mv 2 represent the total momenta at times t x and t 2 respectively. 

CONSTRAINTS. HOLONOMIC AND NON-HOLONOMIC CONSTRAINTS 

7.20. In each of the following cases state whether the constraint is holonomic or non- 
holonomic and give a reason for your answer: (a) a bead moving on a circular wire; 

(b) a particle sliding down an inclined plane under the influence of gravity; (c) a 
particle sliding down a sphere from a point near the top under the influence of gravity. 

(a) The constraint is holonomic since the bead, which can be considered a particle, is constrained 
to move on the circular wire. 

(6) The constraint is holonomic since the particle is constrained to move along a surface which is 
in this case a plane. 

(c) The constraint is non-holonomic since the particle after reaching a certain location on the 
sphere will leave the sphere. 

Another way of seeing this is to note that if r is the position vector of the particle 
relative to the center of the sphere as origin and a is the radius of the sphere, then the 
particle moves so that r 2 § a 2 . This is a non-holonomic constraint since it is not of the 
form (26), page 170. An example of a holonomic constraint would be r 2 = a 2 . 

STATICS. PRINCIPLE OF VIRTUAL WORK. STABILITY 

7.21. Prove the principle of virtual work, Theorem 7.14, page 171. 

For equilibrium, the net resultant force F v on each particle must be zero, so that 

2F„'8r„ = (-0 

V 

But since F„ = F (a) + F (c) where F^ a) and F< c) are the actual and constraint forces acting on the 
pth particle, (1) can be written 

2F£ a) -8r„ + 2Ft c), 8'v = ° (2) 

If we assume that the virtual work of the constraint forces is zero, the second sum on the left of 

(2) is zero, so that we have _ , ^ / a \ 

2 F Co) • dr v = W 

v v 

which is the principle of virtual work. 

7.22. Two particles of masses mi and m 2 are located on a frictionless double incline and 
connected by an inextensible massless string passing over a smooth peg [see Fig. 7-9 
below]. Use the principle of virtual work to show that for equilibrium we must have 

sin a x _ m2 
sin a 2 mi 

where a x and a 2 are the angles of the incline. 



CHAP. 7] 



SYSTEMS OF PARTICLES 



181 



Method 1. 

Let r x and r 2 be the respective position vectors 
of masses m 1 and m 2 relative to O. 

The actual forces (due to gravity) acting on 
m 1 and m 2 are respectively 



F< a) = m xS , 



F< a) = m 2 g 



According to the principle of virtual work, 
2Fj 0) -«r v - 

or F'< 0) • Sr x + F 2 a) • Sr 2 = 



00 



(*) 



A, 




m£ 


^\B 


S\ a i 


Wlj 











Fig. 7-9 



where 5r x and Sr 2 are the virtual displacements of m 1 and m 2 down the incline. Using (1) in (2), 

™i£ • Sr x + wi 2 g • Sr 2 = (3) 

or m x g 8r t sin a x + m 2 g 8r 2 sin a 2 = (4) 

Then since the string is inextensible, i.e. 8r t + 8r 2 = or 8r 2 = —Sri, (4) becomes 

(mrf sin a x — m^ sin a 2 )S^i = 
But since 8r x is arbitrary, we must have m x g sin « x — m 2 g sin a 2 = 0> i-e»> 

sin ax _ m 2 
h 



(5) 



sm a 2 w 

Method 2. 

When it is not clear which forces are constraint forces doing no work, we can take into account 
all forces and then apply the principle of virtual work. Thus, for example, taking into account 
the reaction forces R x and R 2 due to the inclines on the particles and the tension forces T x and T 2 , 
the principle of virtual work becomes 



(m x g + T x + Rj) • 8r x + (m 2 g + T 2 + R 2 ) • Sr 2 = 

Now since the inclines are assumed smooth [so that 
the reaction forces are perpendicular to the in- 
clines] we have 

Ri'Srj = 0, R 2 *8r 2 = (7) 

Also, since there is no friction at the peg, the ten- 
sions Tj and T 2 have the same magnitude. Thus 
we have, using the fact that 8r x and Sr 2 are directed 
down the corresponding inclines and the fact that 

8r 2 = -8r lt 

T a • 8r x + T 2 • Sr 2 = -T x 8r x - T 2 8r 2 

= {T 2 -T x )8r x = (8) 
since T x = T 2 . Then using (7) and (5), (6) becomes 

m x g • 5r x + m 2 g • Sr 2 = 
as obtained in (3). Fig. 7-10 



<*) 




7.23. Use Theorem 7.15, page 171, to solve Problem 7.22. 

Let the string have length I and suppose that the lengths of string OA and OB on the inclines 
[Fig. 7-9] are x and l — x respectively. The total potential energy using a horizontal plane 
through O as reference level is 



V = 
Then for equilibrium we must have 



-m x gx sin a t — m 2 g(l — x) sin a 2 



— = —m x g sin a x + m^g sm a 2 = 



sinaj 
sin a 2 



m 2 



It should be noted that V is not a minimum in this case so that the equilibrium is not stable, 
as is also evident physically. 



182 SYSTEMS OF PARTICLES [CHAP. 7 

D'ALEMBERTS PRINCIPLE 

7.24. Use D'Alembert's principle to describe the motion of the masses in Problem 7.22. 

We introduce the reversed effective forces m x r x and m 2 V 2 in equation (3) of Problem 7.22 to 

obtain .. .. 

(m x g- m x r x )'8r x + (m 2 g - ra 2 r 2 ) • Sr a = (1) 

This can be written #> 

(TOjfli sin**! — m x r x )8r x + (ra 2 sina 2 — m 2 'r 2 )Sr 2 = (2) 

Now since the string is inextensible so that r x + r % — constant, we have 

8r x + 8r 2 = 0, r x + V 2 = 
or Sr 2 = — Srj, r 2 = — rV Thus (2) becomes, after dividing by 8r x # 0, 

wijflf sinaj — m^ — m 2 flr sina 2 — m 2 r x = 
TOjjf sin a x — m^ sin a 2 



n = 



m x + m 2 



Thus particle 1 goes down or up the incline with constant acceleration according as 
m x g sin a x > m<$ sin a 2 or m x g sin a x < m^g sin a 2 respectively. Particle 2 in these cases goes 
up or down respectively with the same constant acceleration. 

We can also use a method analogous to the second method of Problem 7.22. 



MISCELLANEOUS PROBLEMS 

7.25. Two particles having masses mi and m 2 move 
so that their relative velocity is v and the veloc- 
ity of their center of mass is v. If M = m\ + w 2 
is the total mass and ju, = miw 2 /(wi + ra 2 ) is the 
reduced mass of the system, prove that the total 
kinetic energy is $Mv 2 + $ixv 2 . 

Let r x , r 2 and f be the position vectors with re- 
spect to O of mass m x , mass m 2 and the center of 
mass C respectively. 

From the definition of the center of mass, we have 

m 1 T 1 + m 2 r 2 
t = and * — i 

r m x + m 2 ™i + ™2 

or using v x = i x , v 2 = r 2 , v = f , _ 

m x \ x + m 2 \ 2 = (m x + m 2 )v K 1 ) 

If the velocity of m x relative to m 2 is v, then 




v = ^(r!-r 2 ) = h - f 2 = v x - v 2 
so that v i - v 2 = v 

Solving (1) and (2) simultaneously, we find 

m 2 v _ _ _ m i v 

v i = V + «, + «,' v 2 - v Wi + m2 

Then the total kinetic energy is 

T = l^ivf + 2 m 2V 2 



m 2 v \ 2 l /_ m i v 

^1% ... _ 1 a^o , 1 



1 / w 2 v \ , 1 /_ '"'' 



(*) 



= !<-+«*■ +1^- = i w + i 



-/iV 2 



CHAP. 7] 



SYSTEMS OF PARTICLES 



183 



7.26. Find the centroid of a uniform semicircular 
wire of radius a. 

By symmetry [see Fig. 7-12] the centroid of 
the wire must be on the y axis, so that x = 0. If o- 
is the mass per unit length of the wire, then if ds 
represents an element of arc, we have ds = ado 
so that 



I y a ds i 

J ads 
r 



(a sin e)(a de) 



c 

2a? 

■n-a 



f- 



de 




2a 



Fig. 7-12 



7.27. Suppose that n systems of particles be given having centroids at f i, f 2 , . . . , f» and 
total masses M x , M 2 , ... .,M n respectively. Prove that the centroid of all the systems 

is a t 

Mif l + M 2 i 2 + • • • + M n r n 



Mi + M 2 + 



+ M n 



Let system 1 be composed of masses m n , m 12 , . . . located at r n , r 12 , . . . respectively. Similarly 
let system 2 be composed of masses m 21 , m 22 , . . . located at r 21 , r 22 , .... Then by definition, 



m n r n 


+ 


m vFn 


+ ••• 


m n 


+ 


m 12 + 




™21*21 


+ 


w 22 r 22 


+ ••• 


m 21 


+ 


m 22 + 




m nl r nl 


+ 


m n2 r n2 


+ '•• 



*2 



™„i + in n2 + 

But the centroid for all systems is located at 

(m n r u + m 12 r 12 +•••) + (m 21 r 21 + m 22 r 22 + 



ra u r u 


+ 


^12*12 


+ ••• 






Mi 




m 21 r 21 


+ 


w 22 r 22 


+ ••• 






M 2 




^nl^nl 


+ 


m n2fn2 


+ ••• 



M m 



■ ) + • • • + (m nl r Ml + m n2 r n2 + • • • ) 



(w u + m 12 + • • ■) + (w 21 + m 22 + • • •) + • • • + (m nl + m n2 H ) 

Mjt x + M 2 r 2 + • • • + M n r n 



M x + M 2 + • • • + Af „ 



7.28. Find the centroid of a solid of constant density 
consisting of a cylinder of radius a and height H 
surmounted by a hemisphere of radius a [see 
Fig. 7-13]. 

Let f be the distance of the centroid of the solid from the 
base. The centroid of the hemisphere of radius a is at dis- 
tance fa + H from the base of the solid, and its mass is 
M x = |^a 3 £r [see Problem 7.11]. 

The centroid of the cylinder of radius a and height H 
is at distance ±H from the base of the solid and its mass 
is M 2 = ira 2 Ho. 

Then by Problem 7.27, 

(|,ra3q)(fa + ff) + {TraWo){%H) 
~ |ira 3 a + 7ro 2 Hff ~ 

- 3«2 + 8aH + 6m 
~ 8a + 12H 



Af, 





/ ' T ■ "V 


» 






*r 




>s>s> *-~. ' ''iliiip' 

4 /"» 






r 


1 ■# 


i 


•1 


2 I ! « 

! M 2 






^- Base of s 


olid 



Fig. 7-13 



184 



SYSTEMS OF PARTICLES 



[CHAP. 7 



7.29. A circular hole of radius a/2 is cut out of a circular region of radius a, as shown 
in Fig. 7-14. Find the centroid of the shaded region thus obtained. 
y v 




'x 



y —%ira 2 o 



x a 



Fig. 7-14 



Fig. 7-15 



By symmetry the centroid is located on the x axis, so that y = 0. 

We can replace the circular region of radius a by the mass M 1 — TraPo concentrated at its 
centroid x x = a [Fig. 7-15]. Similarly, we can replace the circular hole of radius a/2 by the 
negative mass M 2 = —\ira % o concentrated at its centroid x 2 = fa. Then the centroid is located 
on the x axis at 

M l x l + M 2 x 2 (Tra 2 a)(a) + (-%ira 2 a)(%a) _ 5 



Mi + M 2 



ira 2 a — \ira % o 



7.30. A uniform rod PQ [see Fig. 7-16] of mass m and length L has its end P resting 
against a smooth vertical wall AB while its other end Q is attached by means of an 
inextensible string OQ of length I to the fixed point on the wall. Assuming that 
the plane of P, Q and O is vertical and perpendicular to the wall, show that 
equilibrium occurs if 

V4L 2 - I 2 



sin a 



lV& 



sm/3 



There is only one actual force, i.e. the weight rag of the 
rod. Other forces acting are the force of the wall on the 
rod and the tension in the string. However, these are con- 
straint forces and can do no work. This can be seen since 
if P were to slide down the wall no work would be done, 
because the wall is frictionless and thus the force due to 
the wall on the rod is perpendicular to the wall. Also if Q 
were to drop, it could only move perpendicular to the 
string at Q. 

Let r be the position vector of the center of mass C 
[in this case also the center of gravity] relative to 0. Also 
let i and j be unit vectors in the horizontal and vertical 
directions respectively so that r = xi + yj. 



From Fig. 7-16, 



OQ = OP + PQ 
OQ = OC + CQ 



V4L 2 - I 2 



Then from (1), on taking the dot product with i, 

OQ«i = OP'i + PQ'i 

Since OP • i = 0, this reduces to 

OQ'i = PQ'i 

or I sin a = L sin /? 




Fig. 7-16 



(1) 
(2) 



CHAP. 7] 



SYSTEMS OF PARTICLES 



185 



Similarly on taking the dot product of both sides of (2) with j, 

OQ'j = OC'j + CQ'j 

or I cos a — y + \L cos /? (4) 

Now a virtual displacement of the center of mass C is given by 

Sr = 8xi + 8yj (5) 

Since mg is the only actual force, the principle of virtual work becomes 

mg • Sr = (6) 

Using (5), this becomes mg 8y = or 8y = (7) 

Now from (S) and (A), we have 

I COS a 8a = L cos (i SfS 
— I sin a 8a = 8y — \L sin S/3 
since I and L are constants and since 8 has the same properties as the differential operator d. 
Since 8y = from (7), these equations become 

ZcosaSa = Lcos/JS/3 (5) 

J sin a So; = lLsin/?S/? (9) 

From (5) and (P), we have on division, 



Now from (3), 

so that 

Thus equation (10) can be written 



sin « _ 1 sin/? 
cos a ~ 2 cos ^ 

sin fi — (l/L) sin a 

cos fi = Vl - (WL 2 ) sin2 a 

iin a 1 Z sin a 



Vl - sin2« 2 VL2 - J2 sin 2 a 

Dividing by sin a and squaring both sides, we find 



and from (11) 
as required. 



sin/? = 



V4L2- 


Z2 


iVs 


V4L2- 


/2 



Ly/Z 



(10) 

(«) 

(12) 

(IS) 



(U) 
(IS) 



7.31. A uniform solid consists of a cylinder of radius -a 
and height H on a hemisphere of radius a, as indi- 
cated in Fig. 7-17. Prove that the solid is in stable 
equilibrium on a horizontal plane if and only if 
a/H > \/2. 

By Problem 7.28 the centroid C is at a distance CB 
from the center B of the hemisphere given by 

„ _ 3q2 + SaH + 6ff 2 _ 6ff2 - 3a 2 
8a + 12H ~ 8a + 12H 

Then the distance of the centroid C above the plane is 
CP = CD + DP = CB cos e + BQ 
6W - 3a2 



8a + 12H 



cos + a 




186 



SYSTEMS OF PARTICLES 



[CHAP. 7 



so that the potential energy (or potential) is 

V 



n* i 6ff 2 - 3a 2 



tp -,.v • dV /3a 2 -6H2\ . 

Equilibrium occurs where — = or M# ( -= — „ ) sin 9 — 0, i.e. = 0. 

Then the equilibrium will be stable if 

'3a 2 - 6# 2 



aW 
do 2 



= M^ 



,=o v 8a + 12H 

i.e. 3a 2 - 6H 2 > or a/H > V2. 



cos 9 



0=0 



M# 



3a 2 - 6ff 2 \ 
8a + 12H 



> 



7.32. A uniform chain has its ends suspended 
from two fixed points at the same hori- 
zontal level. Find an equation for the 
curve in which it hangs. 

Let A and B [Fig. 7-18] be the fixed points. 
An element of the chain of length As is in equilib- 
rium under the tensions of magnitude T and 
T + AT due to the rest of the chain and also 
the weight ag As of the element of chain. Now 
from Fig. 7-18 if the directions of the vectors 
corresponding to T and T + AT make angles 
of 9 and 9 + A9 with the x axis respectively, 
we have as the condition for equilibrium [neg- 
lecting terms of order (A9) 2 and higher], Fig. 7-18 

(T + AT) cos (9 + A9)i + (T + AT) sin (9 + A9)j - (T cos 9 i + T sin 9 j) - ag] As 

or (T + AT) cos {9 + A9) = T cos 9 

(T + AT) sin (9 + A9) - T sin 9 = ag As 



A 






.B 






//T + 






A A e + Ae 






A*l 


f- — 






J^*r 


agAs 






*T ,, 




/, 




J 








i 





= 



(1) 

(2) 



Equation (1) shows that the horizontal component T cos 9 must be a constant, which we shall take 
as T which corresponds to the tension at the lowest point of the chain, where 9 = 0. Thus 

Tcos9 = T n {3) 



From (2) we find on dividing by A9, 

(T + AT) sin(fl + Afl) 



T sin9 



A9 
Taking the limit of both sides of U) as A9 -* 0, we find 

d 



As 

° 9 A~9 



d9 



(T sin 9) = ag 



ds 
d9 



Using («?) to eliminate T, (5) becomes 

d 



j- e (T tan 9) 
T n 



ds 
° 9 T9 



or 



where b = Tjag. Now 

Thus from (7) and (*), 
dx 
d9 

dy 
de 



do 



dx 
de 

dx ds 
ds d9 

dy ds 
ds d9 



ag 



= COS 9, 



sec 2 9 



b sec 2 9 



dy ■ n 

-f- = sm9 
ds 



= (cos 9)(b sec 2 9) 
= (sin 9){b sec 2 9) 



= b sec 9 



= b sec 9 tan 9 



tt) 

(5) 

(6) 
(7) 
(8) 

(9) 
(10) 



CHAP. 7] SYSTEMS OF PARTICLES 187 

Integrating (9) and (10) with respect to $, we find 

05 = 6 In (sec 6 + tan e) + Cj (11) 

y = b sec $ + c 2 (12) 

Let us assume that at the lowest point of the chain, i.e. at e = 0, x = and y = b. Then 
from (11) and (12) we find Cj = 0, c 2 = 0. Thus 

x = b In (sec + tan e) (IS) 

y = b sec ff (14) 

From (15) we have sec 6 + tan = e* /b (15) 

But sec 2 e — tan 2 e = (sec + tan tf)(sec — tan e) = 1 (10) 

Then dividing (1£) by (15), we find 

sec e — tan 6 = e~*/ b (17) 



Adding (15) to (17") and using (1-4), we find 

This curve is called a catenary [from the Latin, meaning chain] 



y = |(ex/b + e -*/b) = 6 cosh | (15) 



Supplementary Problems 



DEGREES OF FREEDOM 



7.33. Determine the number of degrees of freedom in each of the following cases: (a) a particle moving 
on a plane curve; (6) two particles moving on a space curve and having constant distance between 
them; (c) three particles moving in space so that the distance between any two of them is 
always constant. Ans. (a) 1, (b) 1, (c) 6 

7.34. Find the number of degrees of freedom for a rigid body which (a) moves parallel to a fixed 
plane, (6) has two points fixed but can otherwise move freely. Ans. (a) 3, (6) 1 

7.35. Find the number of degrees of freedom for a system consisting of a thin rigid rod which can 
move freely in space and a particle which is constrained to move on the rod. Ans. 4 



CENTER OF MASS AND MOMENTUM OF A SYSTEM OF PARTICLES 

7.36. A quadrilateral ABCD has masses 1, 2, 3 and 4 units located at its vertices A(— 1,-2,2), 
B(S, 2, -1), C(l, —2, 4) and D(3, 1, 2). Find the coordinates of the center of mass. Ans. (2, 0, 2) 

7.37. A system consists of two particles of masses m x and m 2 . Prove that the center of mass of the 
system divides the line joining Wj to m 2 into two segments whose lengths are in the ratio m? to Wj. 

7.38. A bomb dropped from an airplane explodes in midair. Prove that if air resistance is neglected, 
then the center of mass describes a parabola. 

7.39. Three particles of masses 2, 1, 3 respectively have position vectors r t = 5ti — 2t 2 j + (St — 2)k, 
r 2 = (2t - 3)i + (12 - 5t 2 )j + (4 + 6* - 3*3)k, r 3 = (2* - 1)1 + (« 2 + 2)j - tf» k where t is the time. 
Find (a) the velocity of the center of mass at time t = 1 and (6) the total linear momentum 
of the system at t = 1. Ans. (a) 3i — 2j — k, (6) 181 — 12j — 6k 



188 



SYSTEMS OF PARTICLES 



[CHAP. 7 



7.40. Three equal masses are located at the vertices of a triangle. Prove that the center of mass is 
located at the intersection of the medians of the triangle. 



7.41. A uniform plate has the shape of the region bounded by the parabola 
y = x 2 and the line y — H in the xy plane. Find the center of mass. 
Ans. x = 0, y = ^H 

7.42. Find the center of mass of a uniform right circular cone of radius a 
and height H. 

Ans. That point on the axis at distance £i? from the vertex. 

7.43. The shaded region of Fig. 7-19 is a solid spherical cap of height H 
cut off from a uniform solid sphere of radius a. (a) Prove that the 
centroid of the cap is located at a distance f (2a — H) 2 /(Sa — H) from 
the base AB. (b) Discuss the cases H — 0, H — a and H = 2a. 

7.44. Find the center of mass of a uniform plate bounded by 

y = sin x and the x axis. Ans. x — tt/2, y — ir/& 



7.45. Find the center of mass of a rod of length I whose den- 
sity is proportional to the distance from one end O. 
Ans. |Z from end O 

7.46. Find the centroid of a uniform solid bounded by the 
planes Ax + 2y + z = 8, x — 0, y = 0, 2 = 0. 

Ans. f = j^«(i + 2j + 4k) 

7.47. A uniform solid is bounded by the paraboloid of revolu- 
tion x 2 + y 2 = cz and the plane z = H [see Fig. 7-20]. 
Find the centroid. Ans. x = 0, y = 0, z = %H 




Fig. 7-19 




Fig. 7-20 



ANGULAR MOMENTUM AND TORQUE 

7.48. Three particles of masses 2, 3 and 5 move under the influence of a force field so that their 
position vectors relative to a fixed coordinate system are given respectively by r x = 2ti — 3j + t 2 k, 
r 3 = (t + l)i + 3tj - 4k and r 3 = < 2 i — *j + (2t - l)k where t is the time. Find (a) the total 
angular momentum of the system and (b) the total external torque applied to the system, taken 
with respect to the origin. 

Ans. (a) (31 - 12t)i + (6t 2 - lOt - 12)j + (21 + 5* 2 )k 
(b) -12i + (12* - 10)j + lOtk 

7.49. Work Problem 7.48 if the total angular momentum and torque are taken with respect to the 
center of mass. 

7.50. Verify that in (a) Problem 7.48 and (6) Problem 7.49 the total external torque is equal to the 
time rate of change in angular momentum. 

7.51. In Problem 7.48 find (a) the total angular momentum and (b) the total external torque taken 
about a point whose position vector is given by r = ti - 2tj + 3k. Does the total external torque 
equal the time rate of change in angular momentum in this case? Explain. 

7.52. Verify Theorem 7.9, page 169, for the system of particles of Problem 7.48. 

7.53. State and prove a theorem analogous to that of Theorem 7.9, page 169, for the total external 
torque applied to a system. 



7.54. Is the angular momentum conserved in Problem 7.38? Explain. 



CHAP. 7] 



SYSTEMS OF PARTICLES 



189 



WORK, ENERGY AND IMPULSE 

7.55. Find the total work done by the force field of Problem 7.48 in moving the particles from their 
positions at time t — 1 to their positions at time t = 2. Ans. 42 

7.56. Is the work of Problem 7.55 the same as that done on the center of mass assuming all mass 
to be concentrated there? Explain. 

7.57. Find the total kinetic energy of the particles in Problem 7.48 at times (a) t = 1 and (b) t — 2. 
Discuss the connection between your results and the result of Problem 7.55. 

Ans. (a) 72.5, (6) 30.5 

7.58. Find the total linear momentum of the system of particles in Problem 7.48 at times (a) t = 1 and 
(6) t = 2. Ans. (a) 17i + 4j + 14k, (6) 27i + 4j + 18k 

7.59. Find the total impulse applied to the system of Problem 7.48 from t = 1 to t - 2 and discuss 
the connection of your result with Problem 7.58. Ans. lOi + 4k 

7.60. Prove Theorem 7.13, page 170. 

7.61. Verify Theorem 7.13, page 170, for the system of particles in Problem 7.48. 

CONSTRAINTS, STATICS, VIRTUAL WORK, STABILITY AND D'ALEMBERT'S PRINCIPLE 

7.62. In each case state whether the constraint is holonomic or non-holonomic and give a reason for 
your answer: (a) a particle constrained to move under gravity on the inside of a vertical paraboloid 
of revolution whose vertex is downward; (6) a particle sliding on an ellipsoid under the influence 
of gravity; (c) a sphere rolling and possibly sliding down an inclined plane; (d) a sphere rolling 
down an inclined plane parallel to a fixed vertical plane; (e) a particle sliding under gravity on 
the outside of an inverted vertical cone. 

Ans. (a) holonomic, (6) non-holonomic, (c) non-holonomic, (d) holonomic, (e) holonomic 



7.63. A lever ABC [Fig. 7-21] has weights W t and W 2 
at distances a t and a 2 from the fixed support B. 
Using the principle of virtual work, prove that a 
necessary and sufficient condition for equilibrium is 
W t a x - W 2 a 2 . 

7.64. Work Problem 7.63 if one or more additional weights 
are placed on the lever. 

7.65. An inextensible string of negligible mass hanging 
over a smooth peg at B [see Fig. 7-22] connects one mass 
m x on a frictionless inclined plane of angle a to another 
mass m 2 . Using D'Alembert's principle, prove that 
the masses will be in equilibrium if m 2 = m 1 sin a. 

7.66. Work Problem 7.65 if the incline has coefficient of fric- 
tion ft. Ans. m 2 = m^sin a — fi cos a) 



W* 



B 



IE 



Og 



Fig. 7-21 



W* 




Fig. 7-22 



7.67. A ladder AB of mass m has its ends on a smooth wall and floor 
[see Fig. 7-23]. The foot of the ladder is tied by an inextensible 
rope of negligible mass to the base C of the wall so that the 
ladder makes an angle a with the floor. Using the principle 
of virtual work, find the magnitude of the tension in the rope. 
Ans. \mg cot a 

7.68. Work (a) Problem 7.63 and (6) Problem 7.65 by using the po- 
tential energy method. Prove that the equilibrium in each case 
is unstable. 




Fig. 7- 



190 



SYSTEMS OF PARTICLES 



[CHAP. 7 



7.69. A thin uniform rod of length I has its two ends constrained to move on 
the circumference of a smooth vertical circle of radius a [see Fig. 7-24]. 
Determine conditions for equilibrium. 

7.70. Is the equilibrium of the rod of Problem 7.69 stable or not? Explain. 

7.71. A solid hemisphere of radius a is located on a perfectly rough inclined 
plane of angle a. 

(a) Prove that it is in stable equilibrium if a < sin -1 (3/8). 

(b) Are there any other values of a for which equilibrium can occur? 

Which of these, if any, yield stable equilibrium? Fig. 7-24 

7.72. Use D'Alembert's principle to obtain the equations of motion of masses m 1 and w 2 of Problem 7.65. 

7.73. Work Atwood's machine problem [see Problem 7.22, page 180] by using D'Alembert's principle. 

7.74. Use D'Alembert's principle to determine the equations of motion of a simple pendulum. 




MISCELLANEOUS PROBLEMS 

7.75. Prove that the center of mass of a uniform circular arc of radius a and central angle a is 
located on the axis of symmetry at a distance from the center equal to (a sin a)/a. 

7.76. Discuss the cases (a) a = tt/2 and (6) a = w in Problem 7.75. 



7.77. 



7.78. 



7.79. 



7.80. 



7.81. 



7.82. 



7.83. 



7.84. 



7.85. 



A circle of radius a is removed from a uniform circular plate 
of radius b > a, as indicated in Fig. 7-25. If the distance be- 
tween their centers A and B is D, find the center of mass. 
Ans. The point at distance a 2 D/(b 2 - a 2 ) below B. 

Work Problem 7.77 if the circles are replaced by spheres. 
Ans. The point at distance a 3 D/(b 3 - a 3 ) below B. 

Prove that the center of mass does not depend on the origin 
of the coordinate system used. 

Prove that the center of mass of a uniform thin hemispherical 
shell of radius a is located at a distance £a from the center. 




Fig. 7-25 



Let the angular momentum of the moon about the earth be A. Find the angular momentum of a 
system consisting of only the earth and the moon about the center of mass. Assume the masses 
of the earth and moon to be given by M e and M m respectively. Ans. M e A/(M e + M m ) 

Does Theorem 7.13, page 170, apply in case the angular momentum is taken about an arbitrary 
point? Explain. 

In Fig. 7-26, AD, BD and CD are uniform thin rods 
of equal length a and equal weight w. They are 
smoothly hinged at D and have ends A, B and C on a 
smooth horizontal plane. To prevent the motion 
of ends A, B and C, we use an inextensible string 
ABC of negligible mass which is in the form of an 
equilateral triangle. If a weight W is suspended 
from D so that the rods make equal angles a with 
the horizontal plane, prove that the magnitude of 
the tension in the string is %V% (W + 3w) cot a. 

Work Problem 7.83 if the weight W is removed 
from D and suspended from the center of one of 
the rods. 




Fig. 7-26 



Derive an expression for (a) the total angular momentum and (b) the total torque of a system about 
an arbitrary point. 



CHAP. 7] 



SYSTEMS OF PARTICLES 



191 



7.86. Prove that the torque about any point P is equal to the time rate of 
change in angular momentum about P if and only if (a) P is fixed in 
space, (b) P coincides with the center of mass or (c) P is moving with 
a velocity which is in the same direction as the center of mass. 



7.87. Find the centroid of a solid of constant density consisting of a right 
circular cone of radius a and height H surmounted by a hemisphere of 
radius a [see Fig. 7-27]. 

Ans. At height £(a 2 + H 2 )/(2a + H) above O. 

7.88. Work Problem 7.87 if the density of the cone is twice the density of 
the hemisphere. Ans. At height § (a 2 + 2H 2 )/(a + H) above O. 




7.89. A hemisphere of radius a is cut out of a uniform solid cube of side b > 2a [see Fig. 7-28]. Find 
the center of mass of the remaining solid. 






Fig. 7-28 



Fig. 7-29 



Fig. 7-30 



7.90. A uniform chain of 45kgwt is suspended from two fixed supports 15 meters apart. If the sag 
in the middle is 20 cm, find the tension at the supports. Ans. 450 kg wt 

7.91. A chain of length L and constant density a is suspended from two fixed points at the same 
horizontal level. If the sag of the chain at the middle is at a distance D below the horizontal 
line through the fixed points, prove that the tension at the lowest point of the chain is 
a(L2 - 4D 2 )/8D. 

7.92. Three particles of masses m l ,m 2 ,m 3 are located at the vertices of a triangle opposite sides having 
lengths a lt a 2 ,a 3 respectively. Prove that the center of mass lies at the intersection of the angle 
bisectors of the triangle if and only if mj/aj = m 2 /a 2 = m 3 /a 3 . 

7.93. Masses m x and m 2 are on a frictionless circular cylinder connected by an inextensible string of 
negligible mass [see Fig. 7-29]. (a) Using the principle of virtual work, prove that the system 
is in equilibrium if m 1 sin ai = m 2 sin a 2 . (b) Is the equilibrium stable? Explain. 

7.94. Work Problem 7.93 if friction is taken into account. 

7.95. Derive an expression for the total kinetic energy of a system of particles relative to a point 
which may be moving in space. Under what conditions is the expression mathematically 
simplified? Discuss the physical significance of the simplification. 



7.96. Find the center of mass of a uniform plate shown shaded in Fig. 7-30 which is bounded by the 
hypocycloid x 2 ' z + y 2 ' s = a 2 ' 3 and the lines x = 0, y = 0. [Hint. Parametric equations for the 
hypocycloid are x — a cos 3 e, y = a sin 3 e.] Ans. x — y = 256a/315jr 



192 



SYSTEMS OF PARTICLES 



[CHAP. 7 



7.97. Let m 1( m 2 , m 3 be the masses of three particles and v 12 , v 23 , v 13 be their relative velocities. 

(a) Prove that the total kinetic energy of the system about the center of mass is 

,2 



m x m 2 v\ 2 + m 2 m 3 V23 + m 1 m i v'x Z 
m x + m 2 + m 3 



(6) Generalize the result in (a). 



7.98. A chain of variable density is suspended from two fixed points on the same horizontal level. 
Prove that if the density of the chain varies as the horizontal distance from a vertical line 
through its center, then the shape of the chain will be a parabola. 

7.99. Discuss the relationship of Problem 7.98 with the shape of a suspension bridge. 

7.100. A solid consists of a uniform right circular cone of vertex angle a on a uniform hemisphere of 
the same density, as indicated in Fig. 7-31. Prove that the solid can be in stable equilibrium on 
a horizontal plane if and only if a > 60°. 





Fig. 7-31 



Fig. 7-32 



7.101. A uniform solid [see Fig. 7-32] consists of a hemisphere of radius a surmounted by a cube of 
side b symmetrically placed about the center of the hemisphere. Find the condition on a and b 



for stable equilibrium. 



Ans. alb > yWv- 



7.102. Find the centroid of the area bounded by the cycloid 

x = a(e — sin e), y = a(l — cos e) 
and the x axis. Ans. (ira, 5a/6) 

7.103. Prove that if the component of the torque about point P in any direction is zero, then the 
component of angular momentum about P in that direction is conserved if (a) P is a fixed point, 
(6) P coincides with the center of mass or (c) P is a point moving in the same direction as 
the center of mass. 

7.104. In Problem 7.103, is the angular momentum conserved only if (a), (6) or (c) occurs? Explain. 

7.105. Prove that the virtual work due to a force is equal to the sum of the virtual works which cor- 
respond to all components of the force. 

7.106. Prove that it is impossible for one sphere to be in stable equilibrium on a fixed sphere which is 
perfectly rough [i.e. with coefficient of friction p = 1]. Is it possible for equilibrium to occur 
at all? Explain. 

7.107. A uniform solid having the shape of the paraboloid of revolution cz = x 2 + y 2 , c > rests on 
the xy plane, assumed horizontal. Prove that if the height of the paraboloid is H, then the 
equilibrium is stable if and only if H < f c. 



7.108. Work Problem 7.107 if the xy plane is inclined at an angle a with the horizontal. 



CHAP. 7] SYSTEMS OF PARTICLES 193 

7.109. In Fig. 7-33, AC and BC are frictionless wires in a vertical plane making angles of 60° and 30° 
respectively with the horizontal. Two beads of masses 3 gm and 6 gm are located on the wires, 
connected by a thin rod of negligible mass. Prove that the system will be in equilibrium 
when the rod makes an angle with the horizontal given by tan -1 (#y3 )• 




Fig. 7-33 

7.110. Prove each of the following theorems due to Pappus. 

(a) If a closed curve C in a plane is revolved about an axis in the plane which does not intersect 
it, then the volume generated is equal to the area bounded by C multiplied by the distance 
traveled by the centroid of the area. 

(b) If an arc of a plane curve (closed or not) is revolved about an axis in the plane which does 
not intersect it, then the area of the surface generated is equal to the length of the arc 
multiplied by the distance traveled by the centroid of the arc. 

7.111. Use Pappus' theorems to find (a) the centroid of a semicircular plate, (b) the centroid of a semi- 
circular wire, (c) the centroid of a plate in the form of a right triangle, (d) the volume of a cylinder. 

7.112. Find the (a) surface area and (6) volume of the doughnut shaped region obtained by revolving 
a circle of radius a about a line in its plane at a distance b > a from its center. 

Ans. (a) A^ab, (b) l^a^b 



Chapter 8 APPLICATIONS to 

VIBRATING SYSTEMS, 
ROCKETS and COLLISIONS 



VIBRATING SYSTEMS OF PARTICLES 

If two or more particles are connected by springs [or interact with each other in some 
equivalent manner], then the particles will vibrate or oscillate with respect to each other. 

As seen in Chapter 4, a vibrating or oscillating particle such as the simple harmonic 
oscillator or bob of a simple pendulum, has a single frequency of vibration. In the case 
of systems of particles, there is generally more than one frequency of vibration. Such 
frequencies are called normal frequencies. The motions of the particles in these cases are 
often called multiply -periodic vibrations. 

A mode of vibration [i.e. a particular way in which vibration occurs, due to particular 
initial conditions for example] in which only one of the normal frequencies is present is called 
a normal mode of vibration or simply a normal mode. See Problems 8.1-8.3. 

PROBLEMS INVOLVING CHANGING MASS. ROCKETS 

Thus far we have restricted ourselves to motions of particles having constant mass. 
An important class of problems involves changing mass. An example is that of a rocket 
which moves forward by expelling particles of a fuel mixture backward. See Problems 
8.4 and 8.5. 

COLLISIONS OF PARTICLES 

During the course of their motions two or more particles may collide with each other. 
Problems which consider the motions of such particles are called collision or impact 
problems. 

In practice we think of colliding objects, such as spheres, as having elasticity. The 
time during which such objects are in contact is composed of a compression time during 
which slight deformation may take place, and restitution time during which the shape is 
restored. We assume that the spheres are smooth so that forces exerted are along the 
common normal to the spheres through the point of contact [and passing through their 
centers] . 

A collision can be direct or oblique. In a direct collision the direction of motion of both 
spheres is along the common normal at the point of contact both before and after collision. 
A collision which is not direct is called oblique. 

Fundamental in collision problems is the following principle called Newton's collision 
rule which is based on experimental evidence. We shall take it as a postulate. 

Newton's collision rule. Let v 12 and v^ be the relative velocities of the spheres 
along the common normal before and after impact. Then 

V 12 ~ ~~ eV 12 

194 



CHAP. 8] APPLICATIONS TO VIBRATING SYSTEMS, ROCKETS AND COLLISIONS 



195 



The quantity c, called the coefficient of restitution, depends on the materials of which the 
objects are made and is generally taken as a constant between and 1. If e = the 
collision is called perfectly inelastic or briefly inelastic. If € = 1 the collision is called 
perfectly elastic or briefly elastic. 

In the case of perfectly elastic collisions the total kinetic energy before and after impact 
is the same. 



CONTINUOUS SYSTEMS OF PARTICLES 

For some problems the number of particles per unit length, area or volume is so large 
that for all practical purposes the system can be considered as continuous. Examples are a 
vibrating violin string, a vibrating drumhead or membrane, or a sphere rolling down an 
inclined plane. 

The basic laws of Chapter 7 hold for such continuous systems of particles. In applying 
them, however, it is necessary to use integration in place of summation over the number 
of particles and the concept of density. 



THE VIBRATING STRING 

Let us consider an elastic string such as a violin or piano string which is tightly 
stretched between the fixed points x = and x = I of the x axis [see Fig. 8-1]. If the 
string is given some initial displacement [such as, for example, by plucking it] and is then 
released, it will vibrate or oscillate about the equilibrium position. 



x = 



x = I 




x = l 



Fig. 8-1 



Pig. 8-2 



If we let Y(x, t) denote the displacement of any point x of the string from the equilibrium 
position at time t [see Fig. 8-2], then the equation governing the vibrations is given by the 
partial differential equation 

d 2 Y n d 2 Y 



dt 2 



= c< 



dx 2 



where if T is the (constant) tension throughout the string and a is the (constant) density 
[mass per unit length of string], 



= Tfa 



(2) 



The equation (1) holds in case the vibrations are assumed so small that the slope dY/dx 
at any point of the string is much less than one. 



BOUNDARY-VALUE PROBLEMS 

The problem of solving an equation such as (1) subject to various conditions, called 
boundary conditions, is often called a boundary-value problem. An important method 
for solving such problems makes use of Fourier series. 



196 



APPLICATIONS TO VIBRATING SYSTEMS, ROCKETS AND COLLISIONS [CHAP. 8 



FOURIER SERIES 

Under certain conditions [usually satisfied in practice and outlined below] a function 
f(x), defined in the interval y < x < y + 21 and having period 21 outside of this interval, 
has the series expansion 



f\ x ) = o" ^ ( ttn cos ~~ l — sm ~T~ ) 



where the coefficients in the series, called Fourier coefficients, are given by 

,y + 2l 



-j s*y + 2l 



n-irX . 

cos — r - aa? 



# sm 






7- 



dx 



(3) 
(5) 



Such a series is called the Fourier series of /(a;). For many problems y = or —I. 



ODD AND EVEN FUNCTIONS 

If y = — I, certain simplifications can occur in the coefficients (4) and (5) as indicated 
below: 



1. If f(-x) = f(x), 



a n = j I /(a;) cos — j— aaj, 



Z 



bn = 



(6) 



In such case f(x) is called an even function and the Fourier series corresponding to 
f(x) has only cosine terms. 



2. If f(-x) = -f(x), 



a n = 0, 



6 n 



fX'^> 



sin — =— ax 



{7) 



In such case f(x) is called an odd function and the Fourier series corresponding to f(x) 
has only sine terms. 

If f(x) is neither even nor odd its Fourier series will contain both sine and cosine terms. 

Examples of even functions are x 4 , Sx 6 + 4x 2 — 5, cos a;, e x + e~ x and the function shown 
graphically in Fig. 8-3. Examples of odd functions are x 3 ,2x 5 — 5x 3 + 4, sinx, e x — e~ x and 
the function shown graphically in Fig. 8-4. 

Examples of functions which are neither even nor odd are x 4 + x 3 , x + cos x and the 
function shown graphically in Fig. 8-5. 

fix) 





Fig. 8-3 



Fig. 8-4 



Fig. 8-5 



If a function is defined in the "half period" x = to x = I and is specified as odd, 
then the function is known throughout the interval — I < x < I and so the series which 



CHAP. 8] APPLICATIONS TO VIBRATING SYSTEMS, ROCKETS AND COLLISIONS 



197 



contains only sine terms can be found. This series is often called the half range Fourier 
sine series. Similarly, a function denned from x = to x = I which is specified as even 
has a series expansion called the half range Fourier cosine series. 



CONVERGENCE OF FOURIER SERIES 

Let us assume the following conditions on f(x): 

1. f{x) is defined in y < x < y + 21. H x) 



/(*i + 0) 



f(x) and its derivative f'(x) are piece- 
wise continuous in y < x < y + 21. [A 
function is said to be piecewise contin- 
uous in an interval if the interval can 
be divided into a finite number of sub- 
intervals in each of which the function 
is continuous and bounded, i.e. there 
is a constant B > such that 
—B < f(x) < B. An example of such a 
function is indicated in Fig. 8-6.] 




/(* 2 +0) 



Fig. 8-6 



3. At each point of discontinuity, for example, Xi [or x 2 ] in Fig. 8-6, f(x) has finite limits 
from the right and left denoted respectively by f(xi + 0) and f{xi — 0) [or f(x 2 + 0), 

f(X2~0)}. 

4. f(x) has period 21, i.e. f(x + 21) = f(x). 

These conditions if satisfied are sufficient to guarantee the validity of equation (3) 
[i.e. the series on the right side of (3) actually converges to f(x)) at each point where f(x) 
is continuous. At each point where f(x) is discontinuous, (3) is still valid if f(x) is replaced 
by £[/(# + 0) 4- f(x — 0)], i.e. the mean value of the right and left hand limits. 

The conditions described above are known as Dirichlet conditions. 



Solved Problems 



VIBRATING SYSTEMS OF PARTICLES 

8.1. Two equal masses m are connected by 
springs having equal spring constant k, 
as shown in Fig. 8-7, so that the masses 
are free to slide on a frictionless table 
AB. The walls at A and B to which the 
ends of the springs are attached are fixed. 
Set up the differential equations of 
motion of the masses. 

Let x x i and x 2 i [Fig. 8-8] denote the dis- 
placements of the masses from their equilibrium 
positions C and D at any time t. 



••" nit 

T5WKT ( J— ^RRKP C ,#" 



-OH5ffir> 



Fig. 8-7 



C ni 

^1 *ii u \ x 2 i 



-nmu^- 



"ChJ ^^m^ — £Lj- r ww^- 



P 
Fig. 8-8 



Q 



198 APPLICATIONS TO VIBRATING SYSTEMS, ROCKETS AND COLLISIONS [CHAP. 8 

Consider the forces acting on the first mass at P. There will be a force due to the spring on 
the right given by K (x 2 i — a^i) = K (x 2 - a^i, and a force due to the spring on the left given by 
— KXyi. Thus the net force acting on the first mass at P is 

k(x 2 — x t )i — KX t i 
In the same way the net force acting on the second mass at Q is 

k(x x — x 2 )i — kx 2 i 

Then by Newton's second law we have 

d 2 
m-r^(x x \) — k(x 2 — x 1 )i — KX t i 

d 2 

or m 'x\ = k(x 2 ~ 2Xi) (l) 

m x 2 = k(x 1 — 2x 2 ) (2) 

8.2. Find (a) the normal frequencies and (b) the normal modes of vibration for the 
system in Problem 8.1. 

(a) Let x 1 = A t cos at, x 2 = A 2 cosut in equations (1) and (2) of Problem 8.1. Then we find, 
after simplifying, 

(2K-mu 2 )A 1 - K A 2 = (1) 

-icA 1 + (2/c-m» 2 )A 2 = (2) 

Now if Ay and A 2 are not both zero, we must have 

2k — ffllO 2 — K 

2 = ° <*> 

— jc 2k — Ww 2 

(2/c — ?ww 2 )(2»c — mw 2 ) — k 2 = or m 2 w 4 — 4/cra<o 2 + 3/c 2 = 



Aktyi ± \/16k 2 w 2 — 12/c 2 m 2 
Solving for w 2 , we find <o 2 = — = giving 

(o 2 = /c/m and w 2 = 3/c/m (.4) 

Then the normal (or natural) frequencies of the system are given by 



» 1 / k . 1 / 3/c 

/ = 7T \\~ and / = — \\ — 

2v \ m 2tt \ m 



(5) 



The normal frequencies are also called characteristic frequencies and the determinant (5) 
is called the characteristic determinant or secular determinant. 

(b) To find the normal mode corresponding to u = yficfm, let w 2 = ic/m in equations (i) and (2). 
Then we find 

A t = A 2 

In this case the normal mode of vibration corresponds to the motion of the masses in the same 
direction [i.e. both to the right and both to the left] as indicated in Fig. 8-9. 



Normal mode corresponding to w = yj nlm Normal mode corresponding to w = yZic/m 

Fig. 8-9 Fig. 8-10 

Similarly to find the normal mode corresponding to w = ySK/m, let « 2 = 3ic/m in equations 

(1) and («). Then we find 

A x — —A 2 



CHAP. 8] APPLICATIONS TO VIBRATING SYSTEMS, ROCKETS AND COLLISIONS 199 

In this case the normal mode of vibration corresponds to the motion of the masses in opposite 
directions [i.e. when one moves to the right the other moves to the left, and vice versa] as 
indicated in Fig. 8-10 above. 

In working this problem we could also just as well have assumed, x t = B x sin at, 
x 2 = B 2 sin a t or x x = A x cos at + B x sin at, x 2 = A 2 cos at + B 2 sin at or x x = C x e™*, 
x 2 - C 2 e ibit . 

8.3. Suppose that in Problem 8.1 the first mass is held at its equilibrium position while 
the second mass is given a displacement of magnitude a > to the right of its 
equilibrium position. The masses are then released. Find the position of each mass 
at any later time. 

Writing Wl = ^f^Jm and « 2 = V^c/w, the general motion of both masses is described by 

x x - C 1 cosu 1 t + C 2 sin ujt + C 3 cos<o 2 £ + C 4 sin w 2 i (-0 

x 2 = D x coswx* + D 2 sinwjt + D 3 cos u 2 * + D± sin a 2 t (2) 

where the coefficients are all constants. Substituting these into equation (1) or equation (2) of 

Problem 8.1 [both give the same results], we find on equating corresponding coefficients of cos Wl t, 

sin a^t, cos u 2 t, sin a 2 t respectively, 

Dj = C lf D 2 = C 2 , D 3 = -C 3 , D 4 = -C 4 

Thus equations (1) and (2) can be written 

x x — C x cos o^t + C 2 sin u x t + C 3 cos <o 2 < + C 4 sin a 2 t (3) 

x 2 = Cxcosuit + C 2 sin u x t — C 3 cosu 2 t ~ C 4 sina 2 t (4) 

We now determine Cj,C 2 , C 3 , C 4 subject to the initial conditions 

Xl = 0, x 2 = a, x x - 0, x 2 — at t = (5) 

From these conditions we find respectively 

C 1 + C 3 = 0, C x - C 3 = a, C 2 W! + C 4 <o 2 = 0, C 2 o^ - C 4 w 2 = 

From these we find C x = \a, C 2 = 0, C 3 = -ia, C 4 = («) 

Thus equations (5) and (4) give the required equations 

x i — ^«(cos w x t — cos o> 2 t) (7) 

x 2 — ^a(cos u^t + cos o) 2 t) {8) 

where a x = V '/c/w, w 2 = v 3*/m. 

Note that in the motion described by (7) and (8), both normal frequencies are present. These 
equations show that the general motion is a superposition of the normal modes. This is sometimes 
called the superposition principle. 



CHANGING MASS. ROCKETS 

8.4. Derive an equation for the motion of a v + v v + Av 

rocket moving in a straight line. 



o 1 , o — 



—Aw m + Am 



Let m be the total mass of the rocket at q 

time t. At a later time t + At suppose that 
the mass is m + Am due to expulsion of a mass 
—Am of gas through the back of the rocket. 

Note that —Am is actually a positive quantity Fig. 8-11 

since Am is assumed negative. 

Let v and v + Av be the velocities of the rocket at times t and t + At relative to an inertial 
system with origin at O. The velocity of the mass of gas ejected from the back of the rocket relative 
to O is v + v where — v is the velocity of the gas relative to the rocket. 



200 



APPLICATIONS TO VIBRATING SYSTEMS, ROCKETS AND COLLISIONS [CHAP. 8 

Since the change in momentum of the system is equal to the impulse, we have 
Total momentum at t + At — total momentum at t = impulse 

{(ra + Am)(v + Av) + (-Am)(v + v )} - mv = F At (l) 

where F is the net external force acting on the rocket. 
Equation (1) can be written as 

Av Am Av 

m M ~ V °^ + AT Am = F 

Then taking the limit as At -> 0, we find 

dv dm 

m di ~ v °-dT = F W 



Writing v = vi, v = -v i, F = Fi, this becomes 



dv . dm _ 

m Tt + v ° It = F W 



8.5. Find the velocity of the rocket of Problem 8.4 assuming that gas is ejected at a 
constant rate and at constant velocity with respect to it and that it moves vertically 
upward in a constant gravitational field. 

If the gas is ejected at constant rate a > 0, then m = m -at where m is the mass of the 
rocket at t - 0. Since F = -mg\ (or F - -mg) and dm/dt = -a, equation (S) of Problem 8.4 
can be written 

/ s\ dv . dv aV o 

(m -at) — - a v = -(m - a t)g or ^ = - 9 + ——^ {1) 

Integrating, we find v = -gt - v In (ra - at) + c x (2) 

If v = at t = 0, i.e. if the rocket starts from rest, then 

= - v In m + c x or c x = v In m 

Thus (2) becomes v - —gt + v n \n ( — — — J 

\m - at J 

which is the speed at any time. The velocity is v = vi. 

Note that we must have m - at > 0, otherwise there will be no gas expelled from the rocket, 
in which case the rocket will be out of fuel. 



COLLISIONS OF PARTICLES 

8.6. Two masses mi and m 2 traveling in the same straight line collide. Find the velocities 
of the particles after collision in terms of the velocities before collision. 

Assume that the straight line is taken to be 
the x axis and that the velocities of the particles 
before and after collisions are v,,v 9 and vi, v, 
respectively. 1 /~ \ v * » -v- v v 2 ^ 

By Newton's collision rule, page 194, 

▼i-v£ = e(v 2 - Vl ) (1) Fig. 8-12 

By the principle of conservation of momentum, 

Total momentum after collision = total momentum before collision 

m x \[ + m 2 V2 = m 1 \ 1 + m 2 v 2 (2) 



CHAP. 8] APPLICATIONS TO VIBRATING SYSTEMS, ROCKETS AND COLLISIONS 



201 



Solving (1) and (2) simultaneously, 



V2 = 



(m x - em 2 )v 1 + m 2 (l + e)v 2 

fltj + W&2 

m 1 (l + e)vj + (m 2 — «w 1 )v 2 
m t + m 2 



(S) 



8.7. Discuss Problem 8.6 for the case of (a) a perfectly inelastic collision, (b) a perfectly 
elastic collision. 

(a) Here we put e = in (3) and (4) of Problem 8.6 to obtain 



vi 



m^Vi + m 2 v 2 
Wj + ra 2 ' 



v 2 = 



m^ + m2V 2 

mj + m 2 



Thus after collision the two particles move with the same velocity, i.e. they move as if they 
were stuck together as a single particle. 

(b) Here we put e — 1 in (3) and (-4) of Problem 8.6 to obtain 

(ra x — m 2 )v 1 + 2m 2 v 2 , 2m l \ l + (mj — m^v 2 

v i = ^ j,^ » v 2 = - 



wii + m 2 
These velocities are not the same. 



m, + w 2 



8.8. Show that for a perfectly elastic collision of the particles of Problem 8.6 the total 
kinetic energy before collision equals the total kinetic energy after collision. 

Using the result of Problem 8.7(&), we have 

Total kinetic energy after collision = n w i v i 2 + ~m 2 v 2 2 



= 2 mi 



(m x 



m 2 )v! + 2w 2 v 2 l 2 



1 2 , 1 z 



m x + m 2 

2 



+ 2% 



C2m l v 1 + (w 2 — m!)v 2 ] 2 



= total kinetic energy before collision 



8.9. 



Two spheres of masses mi and ra 2 respectively, collide obliquely. Find their velocities 
after impact in terms of their velocities before impact. 

Let v 1 ,v 2 and vi,v 2 be the velocities of the 
spheres before and after impact respectively, as 
indicated in Fig. 8-13. Choose a coordinate system 
so that the xy plane is the plane of v t and v 2 , and 
so that at the instant of impact the x axis passes 
through the centers C t and C 2 of the spheres. 

By the conservation of momentum, we have 



m 1 v 1 + m 2 v 2 = w^vi + wi 2 v 2 
From Fig. 8-13 we see that 



(1) 




Fig. 8-13 



v i = ^iCcos *i i — sin e x j) 
v 2 = v 2 (cos *2 * — sin e % j) 
v i — ^i(cos <Pi i — sin <f> t j) 
v 2 = v 2 (cos 2 i — sin <f> 2 j) 
Substituting equations (2)-(5) in (1) and equating coefficients of i and j, we have 



(3) 
U) 
(5) 



202 



APPLICATIONS TO VIBRATING SYSTEMS, ROCKETS AND COLLISIONS [CHAP. 8 



ra^ cos e i + m 2 v 2 cos B 2 = m^ cos <p x + m 2 v' 2 cos 2 
m 1 v 1 sin e l + m 2 v 2 sin e 2 = m^v'^ sin </>! + m 2 i> 2 sin <p 2 
By Newton's collision rule, we have 

Relative velocity after impact along x axis 

= — e{relative velocity before impact along x axis} 
or vi • i - v 2 • i = -e(v! • i - v 2 • i) 

which on using equations {2)-(5) becomes 

v[ cos 0! — v 2 cos <f> 2 = —e(v l cos t — v 2 cos e 2 ) 
Furthermore, since the tangential velocities before and after impact are equal, 

Vi'j = vi'j 
v 2 *3 = v 2 «j 
or v x sin e l = v[ sin <f> t 

v 2 sin e 2 — v'2 sm 02 
Equation (7) is automatically satisfied by using equations (12) and (13). 
From equations (6) and (9) we find 

(wj — m 2 e)v 1 cos ! + m 2 (l + e)t> 2 cos e 2 



V x COS 0! = 
V 2 COS 2 = 

Then using (12) and (15) we find 



m 1 + m 2 

m 1 (l + e)Vi cos J + (m 2 — wiie)^ cos #2 
m l + m 2 



v i = ^i(cos 0i i — sin X j) 

(m l — m 2 e)v 1 cos X i + ra 2 (l + e)v 2 cos 2 i 
~~ m t + m 2 

v 2 = v 2 (cos 02 i _ sin $2 J) 

m^l + e)i>! cos *! i + (m 2 — m 1 e)v 2 cos 2 i 

— wij + m 2 



t>i sin e x j 



— t> 2 sin e 2 j 



(7) 



(*) 
(») 

(10) 

(J*) 

(18) 



CONTINUOUS SYSTEMS OF PARTICLES 

8.10. Derive the partial differential equation (1), page 195, for the transverse vibrations 
of a vibrating string. 




x + Ax 



Fig. 8-14 

Let us consider the motion of an element of the string of length As, greatly magnified in 
Fig. 8-14. 

The forces acting on the element due to the remainder of the string are given by the tensions, 
as shown in Fig. 8-14, of magnitude T(x) and T(x + A«) at the ends x and x + As of the element. 



CHAP. 8] APPLICATIONS TO VIBRATING SYSTEMS, ROCKETS AND COLLISIONS 203 

The net horizontal force in direction i acting on the element is 

[T(x + Ax) cos e(x + Ax) - T(x) cos e(x)]i (1) 

The net vertical force in direction j acting on the element is 

[T(x + Ax) sin 6(x + Ax) — T(x) sin e(x)]j (2) 

If we assume that the horizontal motion in direction i is negligible, the net force (1) is zero. 
Using the fact that the acceleration of the element is d 2 Y/dt 2 approximately and that its mass is 
a As where a is the mass per unit length, we have from (2) and Newton's second law, 

d 2 Y 
a As -TT^- j = [T(x + Ax) sin 6 (x + Ax) — T(x) sin e(x)] j (3) 



or, dividing by Ax j, 

As d*Y _ T(x + Aas) sin 6{x + Ax) - T(x) sin 6(x) 
G Ax dt 2 ~ Ax 

or L | /AY\*d 2 Y _ T(x + Ax) sin e(x + As) - T(x) sin e(x) 

a \ \Ax J dt 2 Ax 

Taking the limit as Ax -* 0, this becomes 

' "d 2 Y d 



U) 



•>Rs) 



Since sin 9 = 



U 2 - dx( Tsin ^ & 

tan e dY/dx 



Vl + tan2 e Vl + (dY/dx) 2 

equation (5) can be written 



I + (&Y\ 2 PY = _j_ f TdY/dx__ 
\ \8xJ dt 2 dx 1 Vl + (dY/dx)* 



(6) 



To simplify this equation we make the assumption that vibrations are small so that the slope dY/dx 
is small in absolute value compared with 1. Then we can neglect (dY/dx) 2 compared with 1 and 
W becomes 

If we further assume that the tension T is constant throughout the string and that a is also 
constant, (7) becomes 

c2 idr (8) 



d 2 Y 2 § 2 Y 

dt 2 ~ C dx 2 



where c 2 = T/a. Unless otherwise specified, when we deal with the vibrating string we shall 
refer to equation (8). 



8.11. Derive the equation of Problem 8.10 if the string is horizontal and gravity is taken 
into account. 

In this case we must add to the right hand side of equation (3) of Problem 8.10 the force 
on the element due to gravity 

— mg = —a As gj 

The effect of this is to replace equation (8) of Problem 8.10 by 

d 2 Y _ 2 &Y _ 
dt 2 C dx 2 9 



FOURIER SERIES 

8.12. Graph each of the following functions. 

f 3 <x < 5 
<«)/<*> = |_3 - 5 <*<0 Peri °d = 10 



204 



APPLICATIONS TO VIBRATING SYSTEMS, ROCKETS AND COLLISIONS [CHAP. 8 



/(*) 













Per 
3 


iod 


~*~ 





















1 

-25 


1 1 
-20 -15 


- 


1 1 
-10 -5 





* 
3 

♦ 


1 
5 


-- 


1 1 
10 15 


- 


- 


1 1 
20 25 











Fig. 8-15 

Since the period is 10, that portion of the graph in — 5 < x < 5 (indicated heavy in 
Fig. 8-15 above) is extended periodically outside this range (indicated dashed). Note that f(x) 
is not defined at x = 0, 5, —5, 10, —10, 15, —15, etc. These values are the discontinuities of f(x). 



(b) f{x) 



sin x ^ x ^ -nr 
77 < x < 2tt 



Period = 2tt 



/<*) 





• \ 
/ \ 
/ . \. . . - 




. J 






s 

\ 




/ »- 


s 

/ 

j 


-St 


—2w —r 





V 


2r 3a- 


Av 



Fig. 8-16 

Refer to Fig. 8-16 above. Note that f(x) is defined for all x and is continuous everywhere. 



8.13. Prove 1 sin — y- 

s 



dx 



■ r 



cos — j- dx 



. kirX j 
sin —j— ax 



kirX , 
cos — j— dx 



I kirX 

■j- cos—j- 
kir I 



I . kvx 
kir I 



= if fc= 1,2,3, 



-j — cos kir + j— cos (— kir) 
Kir kit 



= 



7— sin kir — t~ sin (— kir) 
kir kir 



= 



8.14. Prove 



. r l m-n 

(a) 1 cos — j 

(b) f ' si 



m-rrX fltrX , C 

cos—j—dx — \ 



. m-rrX . n-irx , 
sin — i— sin — i— dx 



m¥= n 

1 m = n 



m-rrX TlirX , n 

sin — 5— cos —j- dx = 



where m and n can assume any of the values 1, 2, 3, ... . 

(a) From trigonometry: cos A cos B = £{cos (A - B) + cos (A + B)}, sin A sin B = ^{cos (A - B) - 
cos (A + B)}. 

Then, if m ¥= n, by Problem 8.13, 






Similarly if m ¥= n, 



CHAP. 8] APPLICATIONS TO VIBRATING SYSTEMS, ROCKETS AND COLLISIONS 205 

If m = n, we have 

j^ COS— COS — dx = -J^l+COS—jd* = I 

J. mirx . nwx , 1 C /.. 2mrx\ . 

sm- r sm- r dx = gj (l-cos— )d* = I 



-i • * - ■'-i \ * / 

Note that if m = « = these integrals are equal to 2/ and respectively. 

(6) We have sin A cosB = ^{sin(A-B) + sin(A+B)}. Then by Problem 8.13, if m + n, 
C . mirx nirx , If J . (m — n)vx , . (m + n)irx l , 

J_ t sin _~ cos -r dx = 2 J_ t \ sin — r— + *™- — i 1 -)** = o 

If m = n, rl i 

( o,-~ m ^ x „~„ nvX j 1 I 2mrx , 

I sin — =— cos -y- dx = k I sm — i — do; = 

The results of parts (a) and (6) remain valid even when the limits of integration — I, I are 
replaced by y, y + 21 respectively. 



8.15. If «, , N 

f[x) = A + 2, ( «nCos— y h b n sm— J— J 

prove that by making suitable assumptions concerning term by term integration of 
infinite series, that for n = 1, 2, 3, . . . , 

(a) a n = \^ f{x)co^dx f (b) b n = \ f f(x)sin^dx, (c) A = ^. 

(a) Multiplying 

f( x) = A + J^^cos^f + 6 n sin^ (i) 

by cos -|- and integrating from -Z to I, using Problem 8.14, we have 

J%(*)coB*p«fe = A ^ cos^f da (f) 

= a m l if m¥= 



Thus 



~ 7 J '^ cos— |— da; if m = 1, 2, 3, . . . 

(6) Multiplying (1) by sin^p and integrating from -I to I, using Problem 8.14, we have 

£/(«). in Spcfc = A J' sin^fd* (Jf) 

+ Si {«*£ sin^cos^fd, + 6„£ sin^sin^f, 



— 6 m Z 



206 APPLICATIONS TO VIBRATING SYSTEMS, ROCKETS AND COLLISIONS [CHAP. 8 

Thus b m = j I f(x) sin— j^-dx if m = 1,2, 3, ... 

(c) Integration of (1) from — I to Z, using Problem 8.13, gives 

f /(«)<*» = 2AI or A = ^J f(x) dx 

If 1 _ «o 

Putting m = in the result of part (a), we find o = t f(x) dx and so A — -~ . 

The above results also hold when the integration limits —I, I are replaced by y, y + 21. 

Note that in all parts above we have assumed interchange of summation and integration. 
Even when this assumption is not warranted, the coefficients a m and b m as obtained above are 
called Fourier coefficients corresponding to f(x), and the corresponding series with these values 
of a m and b m is called the Fourier series corresponding to f(x). An important problem in this 
case is to investigate conditions under which this series actually converges to f(x). Sufficient 
conditions for this convergence are the Dirichlet conditions given on page 197. 

8.16. (a) Find the Fourier coefficients corresponding to the function 

TO -5<z<0 
f(x) = i Period = 10 

n ' [3 0<z<5 

(b) Write the corresponding Fourier series. 

(c) How should f(x) be defined at x = -5, x = and x = 5 in order that the Fourier 
series will converge to f(x) for -5 ^ x ^ 5? 

The graph of f(x) is shown in Fig. 8-17 below. 

I 
— Period ■- 



-10 



— r 

10 



Fig. 8-17 



(a) Period = 21 - 10 and I - 5. Choose the interval y to y + 21 as -5 to 5, so that y = —5. Then 

a n = \f + 2l f(x)cos^dx = |J 5 /(*)cos^ 

y _5 

= | { f (0) cos^f dx + Jf * (3) cos^f «fa| = | jT cos^ dx 

= f^sin^f = if n#0 

5 V n^r o 



o 

'5 n .. o /-5 



3 r 5 ovx , 3 r 5 , _ 

If n = 0, a n = a = -r 1 cos -=- da; = g I da; = 3. 

b n - yj /(*) sm-y- das = gj /(as) sin -g- das 

7 —5 



3 / 5 mrx\ I 5 _ 3(1 - cos rnr) 

-I cos > l — 



5 \ nir 5 / 



CHAP. 8] APPLICATIONS TO VIBRATING SYSTEMS, ROCKETS AND COLLISIONS 207 

(b) The corresponding Fourier series is 

a ° i v< ( ~ nirX i i. • nirx\ 3 , ~ 3(1 — cos nv) . nwx 
- + ^ («n cos — + b n sm— j = 2 + B 2 ^ sm-g- 

3 , 6 / . irx . 1 . 3tr« . 1 . 5vx , 
= 2 + ^( Sin -5- + 3 Sin -5- + 5 Sm -T + ' 

(c) Since f(x) satisfies the Dirichlet conditions, we can say that the series converges to f(x) at all 

points of continuity and to " x + °^ " x ~ °^ at points of discontinuity. At * = -5, and 5, 

which are points of discontinuity, the series converges to (3 + 0)/2 = 3/2 as seen from 
the graph. If we redefine f(x) as follows, 

'3/2 x = -5 

-5 < x < 

f(x) = < 3/2 * = Period = 10 

3 < x < 5 

.3/2 x = 5 

then the series will converge to f(x) for — 5 ^ x ^ 5. 

8.17. If f(x) is even, show that (a) a n = | J" f(x) cos^dx, (b) b n = 0. 

(a) «» = J f /(«) cos^ cfe - f J"° /(») cos ^ ^ + I J' /(*) cos^ dx 

Letting x = —u, 

f /"/<*) cos*|*<fc = ljV*)cos(^)d % = lJ*V)cos^fd M 
since by definition of an even function f(—u) = f(u). Then 

«* = y//(«)coB^d« +|J , /(*)eos^«te - fff(x)cos^dx 

W *n = 7 £ /(^)sin^f d* = |J° /(*)sin^<*s + \f f(x)sm?fLdx (1) 

If we make the transformation x = -u in the first integral on the right of (I), we obtain 
\ J° fix) sin 2p ifa = | J* f{-u) sin (=f^) <fc* = -] f f(- u ) sin ^ <* M 

= -j f f(u)Bin^du = -\ f l f(x)sin^dx (2) 

where we have used the fact that for an even function f(-u) = f(u) and in the last step that 
the dummy variable of integration u can be replaced by any other symbol, in particular x. Thus 
from (1), using (2), we have 



1 j , M . n-n-x . 1 f l ,. . . mtx , 

~lj f( x > sm —j— dx + l I /(*) sin— z-dx = 



8.18. Expand f(x) - x, < x < 2, in a half range (a) sine series, (b) cosine series. 

(a) Extend the definition of the given function to that of the odd function of period 4 shown in 
Pig. 8-18 below. This is sometimes called the odd extension of /(»). Then 21 — 4, 1 = 2. 



208 



APPLICATIONS TO VIBRATING SYSTEMS, ROCKETS AND COLLISIONS [CHAP. 8 



/ 



/ 



+ 



/ 



/ 



/ 



/ 



/(*) 



O 



-1 71 1 7 

-6 /-4 -2 y 

/ ' 

/ 



/ 



/ 



/ 



if 



/ 



/ 



/ 



/ 



/ 



/ 



/ 



/ 



Fig. 8-18 



Thus a n ~ and 
b 



n = 1 J /(») sin —j— dx 



r 



. nirx . 
. x sm-y- dx 

~0 



{w(^-¥)-«(k 



—4 , nirx \ 
—5 sin -g- ) 



cos«jt 



Then 



,, v v< — 4 . %7ra; 

/(a;) — 2i — cos nir sin — — — 
n = 1 %7T 2 

4 / . 77-3 1 . 2ttx , 1 . 3jra; 
- sin T -2 S in^-+ gSin^- 



(b) Extend the definition of f(x) to that of the even function of period 4 shown in Fig. 8-19 below. 
This is the even extension of f(x). Then 21 = 4, 1 = 2. 




Fig. 8-19 



Thus b n = 0, 






,. N nirx , 2 (" 



a; cos— y- dx 



. . / 2 . rura \ M v / —4 

(*) — sm-j- ) — (1) -5- 

\71tt 2 / \n 2 ir 



—4 »Hr£C 



r 2 

If n = 0, a = I 



•M/V 



asda; = 2. 



r (cos 1177 — 1) if n ¥^ 



Then 



/(a?) = 1 + > -^-^(cosrair — 1) cos— r— 



5irx 



wx , 1 SwX 1 „„„, 

2V COS T + 32 COS "g- + g2COS^~ + 



It should be noted that the given function f(x) = x, < x < 2, is represented equally well 
by the two different series in (a) and (6). 



CHAP. 8] APPLICATIONS TO VIBRATING SYSTEMS, ROCKETS AND COLLISIONS 209 

SOLUTIONS OF VIBRATING STRING PROBLEMS 

8.19. Find the transverse displacement of a vibrating string of length I with fixed endpoints 
if the string is initially given a displacement f(x) from its equilibrium position and 
then released. 

Let the transverse displacement of any point x of the string at time t be Y(x, t). Since the 
ends x — and x — I of the string are fixed, we must have Y(0, t) = and Y(l, t) = 0. Since 
the initial displacement is f(x), we have Y(x, 0) = f(x); and since the initial velocity is zero, we 
have Y t (x, 0) = where Y t denotes the partial derivative with respect to t. We must thus solve 
the boundary-value problem 

&X. - o d2Y 

a<2 ~ c dx* w 

Y(0,t) = 0, Y(l,t) = 0, Y(x,0)=f(x), Y t (x,0) = (2) 

Assume a solution to (1) of the form Y = XT where X depends only on x and T depends 
only on t. Then substituting into (1), using X" to denote d 2 X/dx 2 and T" to denote d 2 T/dt 2 , we have 



X T" = c 2 X" T 

X" _ T" 
X ~ c 2 T 



(3) 



Since one side depends only on x and the other side depends on t while x and t are independent, 
the only way in which (2) can be valid is if each side is a constant, which we shall take as 
-X 2 . Thus 

*? - II - _ x2 
X ~ c 2 T ~ 

or X" + \*X = 0, T" + \ z cH = 

These equations have solutions 

X = A x cos \x + Bi sin Xx, T = A 2 cos Xct + B 2 sin Xct 
Thus a solution is given by 

Y(x, t) = XT = {A x cos \x + B x sin \x)(A 2 cos Xct + B 2 sin Xct) (i) 

From the first condition in (2), we have 

B X {A 2 cos Xct + B 2 sin Xct) — 

so that A t = [since if the second factor is zero then the solution is identically zero, which we 
do not want]. Thus 

Y(x, t) = B x sin Xx (A 2 cos Xct + B 2 sin Xct) 

= sin Xx (6 cos Xct + a sin Xct) (5) 

on writing B t A 2 — b, B X B 2 — a. 

Using the second condition of (2 ) in (5), we see that sin \l = or Xl = n-ir where 
n = 1,2,3, ... . Thus X = mr/l and the solution so far is 

v(„ t \ — „• nwx ( , nirct . . nvct\ , , 

x (x, t) — sin —j- I b cos — j — + a sin — j— j (6) 

By differentiating with respect to t, this becomes 

v /„ *\ _ „•», nirX ( nircb . nvct nirca nirct\ 
Y t {x,t) - sin—^ T-sn-T + - r cos- r J 

so that the fourth condition in (2) gives 



Y t (x,0) = S in^(™p) = 



from which a = 0. Thus (6) becomes 



v/~ *\ — u • n7rX n-n-ct 

Y(x, t) — b sin -j- cos — y— (7) 



210 APPLICATIONS TO VIBRATING SYSTEMS, ROCKETS AND COLLISIONS [CHAP. 8 

To satisfy the third condition of (2) we use the fact that solutions of (7) multiplied by constants 
as well as sums of solutions are also solutions [the superposition theorem or principle for linear 
differential equations]. Thus we arrive at the solution 

T7-/ ja v u nvx nvct ,„. 

Y(x, t) = 2i b n sin —j— cos — — (8) 

n=l I I 

Using the third condition of {2) in (8) we must have 

00 

Y(x,0) = f(x) = 2 6„ sin^ (9) 

n=l I 

But this is simply the expansion of f(x) in a Fourier sine series and the coefficients are given by 

b n = T I f( x ) sin— ^— dx 
1 Jo l 

Thus the solution is given by 

Y(x,t) = ^1 j|j* l /(x)sin^^jsin^cos^ (10) 

The method of solution assuming Y — XT is often called the method of separation of variables. 



8.20. A string with fixed ends is picked up at its 
center a distance H from the equilibrium 
position and released. Find the displace- 
ment at any position at any time. 

From Fig. 8-20 we see that the initial displace- 
ment of the string is given by 

(2Hx/l 0<x^l/2 

Y(x,0) = f(x) = \ 2H{l . x)/l |/2 Si * ^ J 




NOW n fl 





b n = j I f(x) sin -j— dx 

f f l/2 2Hx . nirx , , C X 2H . mrx . 

< I — j— sm -j— dx + I —j- (t — sc) sm —r~ dx 



- %K sin ( w "-/ 2 ) 
v 2 n 2 

on using integration by parts to evaluate the integrals. Using this in equation ilO) of Problem 8.19, 

we find 

„. JX SH ^ sin imr/2) . nvx nirct 
Y{x, t) = —z- 2i 9 sm ~T~ cos — r~ 

* 2 n=l n l l 

SH f 1 . -n-X vet 1 . 3rrx Sirct 1 . 5n-aj 5vct 

= ^2\Y 2Sm T c0S -r ~ ¥ sm ~r cos ~r + 52 sin_ r cos ~r " 



8.21. Find the normal frequencies and normal modes for the vibrating string in 
Problem 8.20. 

The normal mode corresponding to the lowest normal frequency is given by the first term in 
the solution of Problem 8.20, i.e., 



8H . »■»_._. vet 
The frequency is given by f x where 



sm —=- cos , 
r 4 l <> 



2Wi = T or A = « ~ «\7 



CHAP. 8] APPLICATIONS TO VIBRATING SYSTEMS, ROCKETS AND COLLISIONS 



211 



Since the cosine varies between —1 and +1, the mode is such that the string oscillates as in 
Fig. 8-21 from the heavy curve to the dashed curve and back again. 




Fig. 8-21 




Fig. 8-22 



The next higher frequency is given by the mode corresponding to the next term in the 
series which, except for sign, is 

81? . 3irx Zirct 
^2 sin— cos-y- 

In this case the frequency is given by 

2W 3 _ — or fs = Yl = 2^- 
The mode is indicated in Fig. 8-22. 

The higher normal frequencies are given by 

f * = 2z\7' fl = 2i\7> '" 
The amplitudes of modes corresponding to the even frequencies 

/a = 2!\a' /4 = 21 \ 7' *" 

are zero, so that these frequencies are not present. In a general displacement, however, they 
would be present. 

Because of the fact that all higher normal frequencies are integer multiples of the lowest 
normal frequency, often called the fundamental frequency, the vibrating string emits a musical 
note. The higher frequencies are sometimes called overtones. 



8.22. Find the transverse displacement of a vibrating string of length I with fixed end- 
points if the string is initially in the equilibrium position and is given a velocity 
distribution defined by g(x). 

In this case we must solve the boundary-value problem 



a 2 r _ 2 <py 

3<2 ~ C 5*2 



CO 



Y(0,t) = 0, Y(l,t)=zO, Y(x,0) = 0, Y t (x,0) = g(x) 



The method of separation of variables and application of the first two conditions of (2\ vields 
as in Problem 8.19, ; 

V(rr +\ — „J„ nirX ( U niTCt i • % "" C * \ 

X \x, t) — sin —j— 6 cos — : — + a sin — =— ) 



I \" wo I "■■ " ai " i J 
However, in this case if we apply the third condition of (2) we find 6 = 0, so that 

w~ *\ _ • nvX . nvct 

Y(x, t) = a sin — j— sin -^~- 

To satisfy the fourth condition we first note that the superposition principle applies, so that we 
arrive at the solution 

x{x, t) — 2, a n sin —f— sin 
»=i ' 



I 



212 APPLICATIONS TO VIBRATING SYSTEMS, ROCKETS AND COLLISIONS [CHAP. 8 

From this we have by differentiation with respect to t, 

v , . N _ £, nirca n _ nwX n ct 
Y t (x, t) — 2i — i — sm ~t~ cos ~ ; — 

n=X l I I 

or Y t (x, 0) = g{x) = 2 — i — sin ~T 

w=l * l 

Then by the method of Fourier series we see that 
mrca n 2 



I 
Thus the required solution is, on using (-4) in 



2 C , \ • n T?% j 2 C . nirx . 

7 I g(x) sm—7— dx or a n ■— I g(x) sm—j— ax 

I J I nirc J I 



<*) 



sin — -. — (o) 



v/ i\ S I 2 f , . . mrx . 1 . nvx . 

Y(x,t) = 2 x |— J o ir(«) ain- r <te| sm— si 

8.23. Find the transverse displacement of a vibrating string of length I with fixed end- 
points if the string initially has a displacement from the equilibrium position given 
by f(x) and velocity distribution given by g(x). 

The solution to the given problem is the sum of the solutions to the Problems 8.19 and 8.22. 
Thus the required solution is 

nirx nirct 
sm — — cos 



Y(x,t) ■= 2 "I7 j /(*) sin^j^- dx 



I WB I 



nirX . nrrct 



+ 2 1 I g(x)-sm-^-dx> sin— r- sm 

n=l «J/ W I I 



MISCELLANEOUS PROBLEMS 

8.24. A particle is dropped vertically on to a fixed horizontal plane. If it hits the plane 
with velocity v, show that it will rebound with velocity — cv. 

The solution to this problem can be obtained from the results of Problem 8.6 by letting m 2 
become infinite and v 2 = while v x = v [where subscripts 1 and 2 refer to the particle and 
plane respectively]. Then the respective velocities after impact are given by 

{{mjm^ — e}v 
lim v( = lim ., , , , — r— = — ev 
m 2 -+°o m 2 -+oo 1 + {m 1 /m 2 ) 

(7n 1 /m 2 )(l + e)v 

lim Vo = lim — ; — r— : 1 — r- = 

Thus the velocity of the particle after impact is -ev. The velocity of the plane of course 
remains zero. 

8.25. Suppose that the particle of Problem 8.24 is dropped from rest at a height H above 
the plane. Prove that the total theoretical distance traveled by the particle before 
coming to rest is given by #(1 + e 2 )/(l - e 2 ). 

Let v be the speed of the particle just before it hits the plane. Then by the conservation of 
energy, \mv 2 + = + mgH or v 2 - 2gH. Thus by Problem 8.24 the particle rebounds with 
speed ev and reaches a height {ev) 2 /2g = e 2 H. It then travels back to the plane through the 
distance e 2 H. Thus on the first rebound it travels through the distance 2e 2 H. 

By similar reasoning we find that on the second, third, . . . rebounds it travels through the 
distance 2e 4 H, 2e 6 H , .... Then the total theoretical distance traveled before coming to rest is 

H + 2e 2 H +2e*H + 2e6#+ ••• = H + 2«*ff(l + e 2 + e* + • • •) = H + ^^ = #(fz"£) 

using the result l + r + r 2 + r s +--- ~ 1/(1 - r) if \r\ < 1. 



CHAP. 8] APPLICATIONS TO VIBRATING SYSTEMS, ROCKETS AND COLLISIONS 



213 



8.26. Two particles having masses m and M 
are traveling on the x axis (assumed 
frictionless) with velocities vd and V\i 
respectively. Suppose that they collide 
and that after the collision (impact) their 
velocities are v 2 i and V 2 i respectively. 
Prove that the velocities of the center of 
mass before and after collision are equal. 

By the conservation of momentum, 

Total momentum before impact 

mt^i + MVxi 

or mVl + MV t 



O 



m 



Fig. 8-23 

total momentum after impact 
mv 2 \ + MV 2 i 
mv 2 + MV 2 



Let x and X be the respective coordinates of the particles. Then the center of mass is given 
by f = (mx + MX)/(m + M). 

The velocity of the center of mass before impact is 

r x — (mv, + MVJ/im + M) 
The velocity of the center of mass after impact is f 2 = (mv 2 + MV 2 )/(m + M). Thus ^ = *f 2 . 



8.27. 



A particle of mass m slides down a frictionless incline of angle <*, mass M and 
length L which is on a horizontal frictionless plane [see Fig. 8-24]. If the particle 
starts initially from rest at the top of the incline, prove that the time for the particle 

to reach the bottom is given by 



J 2L(M + m si n 2 a) 



(M + m)g sin a 



Choose a fixed vertical xy coordinate 
system as represented in Fig. 8-24. Let R be 
the position vector of the center of mass C 
of the incline, A the (constant) vector from 
C to the top of the incline, and s the position 
vector of the particle relative to the top of 
the incline. Then the position vector of par- 
ticle m with respect to the fixed coordinate 
system is R + A + s. Since the only force 
acting on the particle is the weight mg of the 
particle, we have by Newton's second law 
applied to the particle, 

d 2 ,„ . . . . 

mg (1) 



m-jp (R+A + s) = 



dt* + dt 2 




(2) 



Fig. 8-24 



Writing R- Xi+Yj, g - -^-j and s = ss lt where Sl is a unit vector down the incline in the 
direction of s, (2) becomes 



d 2 X . d% 

dt 2 * + dt 2 Sl ~ 



Multiplying by s, • , this becomes 



d 2 X cPs 

dt 2 Sl ' * + dt 2 Sl ' Sl 



-91 



ffSi ' J 



d 2 X d 2 s 

— COSa +__ 



9 sin a 



(») 



214 APPLICATIONS TO VIBRATING SYSTEMS, ROCKETS AND COLLISIONS [CHAP. 8 

Since the net horizontal force acting on the system consisting of the particle and incline is zero, the 
total momentum in the horizontal direction before and after the particle starts sliding is zero. Then 

M^-i + m~CR + A + s)'i = 
at dt 



This can be written as 



(M + m) -j- m -j- cos a = (-4) 



Differentiating (4) with respect to * and solving for d 2 X/dt 2 , we find 

d 2 X _ m cos a cPs 
dt 2 ~ M + m dt 2 

Substituting into (3) yields 

(j2 s (M + m)g sin a (M + m)g sin a 

dt 2 ~~ M + m — m cos 2 a ~ M + m sin 2 a 

Integrating (6) subject to the conditions 8 = 0, ds/dt = at * = 0, we find 

1 C(M + m)g sin a] 



(5) 



(6) 



s = 



t 2 



2 J I + m sin 2 a 
which, when 8 — L, yields the required time. 



8.28. Solve the vibrating string Problem 8.19 if gravity is taken into account. 

The boundary-value problem is 

d 2 Y 2 d 2 Y m 

F(0, t) = 0, Y(l, t) = 0, Y(x, 0) = f(x), Y t (x, 0) = (2) 

Because of the term — g the method of separation of variables does not work in this case. In 
order to remove this term, we let 

Y(x,t) = Z(x,t) + f(x) (3) 

in the equation and conditions. Thus we find 

d 2 Z „d 2 Z , „ „ m 

Z(0, t) + \fr(0) = 0, Z(l, t) + f(l) = 0, Z(x, 0) + f(x) = /(*), Z t (x, 0) = (5) 

The equation (4) and conditions (5) become similar to problems already discussed if we choose $ 
such that 

efy" - g = 0, *(0) = 0, f(l) = (6) 

In this case (4) and (5) become 

dt 2 c dx 2 Kn 

Z(0, t) = 0, Z(l, t) = 0, Z(x, 0) = f{x) - f(x), Z t (x, 0) = (8) 

Now from (6) we find f" - g/c 2 or $(x) = gx 2 /2c 2 + c x x + c 2 ; and since ^(0) = 0, f(l) = 0, 
we obtain c 2 = 0, c± = —gl/2c 2 . Thus 



CHAP. 8] APPLICATIONS TO VIBRATING SYSTEMS, ROCKETS AND COLLISIONS 215 

The solution to equation (7) with conditions (8) is, as in Problem 8.19, 



Z(x, t) = 2 i | f [/(*) - *(*)} sin ^ dx 



. nirx nirct 

sin — — cos 



I I 

and thus 

Y(x, t) = ^2 || f [/<*) - JL ( X 2 _ fa) J sin « da A sin 2^ cos «** + J?L (jB 2 _ lx) 

8.29. Assume that a continuous string, which is fixed at its endpoints and vibrates 
transversely, is replaced by N particles of mass m at equal distances from each other. 
Determine the equations of motion of the particles. 

We assume that the particles are con- 
nected to each other by taut, elastic strings 
having constant tension T [see Fig. 8-25]. 
We also assume that the horizontal distances 
between particles [i.e. in the direction of the 
unit vector i] are equal to a and that the 
transverse displacement [i.e. in the direction 
of the unit vector j] of particle v is Y v . We 
assume that there is no displacement of any 
particle in direction i or — i. Fig. 8-25 

Let us isolate the rfh particle. The forces acting on this particle are those due to the (v - l)st 
and {v + l)st particles. We have 

fY — Y _ \ 
Transverse force due to {v - l)st particle = —T ( — — ) j 

fY — Y \ 
Transverse force due to {v + l)st particle = —T ( — — ) j 




Then by Newton's second law the total transverse force acting on particle v 



Si = -*(*^=0i-r(*^)i 



m d* 

d?Y v T 

m -W = -(Y*-i-2Y v + Y v + 1 ) 

le - Y» = £(^-i-2^+^ + i) (1) 

To take into account the fact that the endpoints are fixed, we assume two particles cor- 
responding to v = and r = N+l for which Y = 0, Y N + 1 = 0. Then on putting „ = 1 and 
v = N in equation (1), we find 

^ = ^- 2Y i + Y ^> ? s = ^(Y N _ 1 -2Y N ) W 



8.30. Obtain the secular determinant condition for the normal frequencies of the system 
of particles in Problem 8.29. 

Let Y v = A v cosat in equations (1) and (2) of Problem 8.29. Then after simplifying we find 

-A„_! + (2- ma a 2 /T)A v - A v + 1 = v = 2, . . ., N- 1 (l) 

(2 - ma«?IT)A x - A 2 = 0, ~A N _ 1 + (2 - ma^/T)A N = (2) 

Putting 2 - mtuP/T = c (3) 

these equations can be written 

cA 1 -A 2 = 0, -A 1 + cA 2 -A 3 = 0, ..., -A N _ 1 + cA N = 



216 



APPLICATIONS TO VIBRATING SYSTEMS, ROCKETS AND COLLISIONS [CHAP. 8 



Then if we wish solutions such that A v ¥= 0, we must require that the 2Vth order determinant 
of the coefficients be zero, i.e., 



A W = 



c 


-1 








.. 











-1 


c 


-1 





.. 














-1 


c 


-1 


.. 























.. 


. -1 


c 


-1 














.. 





-1 


c 



= 



The normal frequencies are obtained by solving this equation for the N values of w 2 . 

Although we have used Y v = A v cos cot, we could just as well have assumed Y v = B v sin wt 
or Y v = A v cos at + B v sin at or Y v = C v e iat . The secular determinant would have come out to be 
the same [compare the remarks at the end of Problem 8.2(6)]. 



8.31. Prove that the normal frequencies in Problem 8.30 are given by 

2T 



<D„ — 



ma 



cos 



N + l 



a= 1, ...,N 



By expanding the determinant A N of Problem 8.30 in terms of the elements in the first row, 
we have 

A N = cA N -i - A N _ 2 (I) 



Also, 



A x = c, A 2 = c 2 - 1 



Putting N = 2 in (i), we see that equations (2) are formally satisfied if we take A = 1. Thus 
conditions consistent with (1) and (2) are 

A = 1, A! = c (3) 

To solve the difference equation (1), assume that A N = p N where p is a constant to be 
determined. Substituting this into (1), we find on dividing by p N ~ 2 , 



p 2 — cp + 1 
If we call c = 2 cos 0, then 



V = 



c ± Vc 2 -4 



,±i0 



p = cos 6 ± t sin e 
Thus solutions of the difference equation are 

( e ifl)N = e iN0 = cos Ne + i sin Nff and (e~ ie ) N = e~ Nie = cos No - i sin No 

Since constants multiplying these solutions and sums of solutions are also solutions [as in 
the case of linear differential equations], we see that the general solution is 

A N = G cos Ne + H sin Ne (4) 

Now from equations (3) we have A = 1, A t = 2 cos e so that G = 1, H = cot e. Thus 

sin Ne cos e sin (N + l)e 



= cos Ne + 



sins 



sine 



This is equal to zero when sin (N + 1)6 = or e = aw/(N +1), a = l,...,N. Thus using 
of Problem 8.30, we find 



2 2T fl «* 

o 2 = 1 — COS t. 7 i ., 

« ma\ N + l 



(5) 



(6) 



CHAP. 8] APPLICATIONS TO VIBRATING SYSTEMS, ROCKETS AND COLLISIONS 217 

8.32. Solve for A v in Problem 8.30 and thus find the transverse displacement Y v of particle v. 

From equations (1) and (2) of Problem 8.30 we have [on using the normal frequencies (6) of 
Problem 8.31 and the superscript a to indicate that the A's depend on a], 

-Ail\ + 2AJ"> cos^^ - A%\ = (1) 

together with the end conditions 

K°° = 0, A^l, = (2) 

The equation (1) subject to conditions (2) can be solved in a manner exactly like that of Problem 8.31, 
and we find 

I (a) - ■ av * 



A^ = C«sin 



N + l 

where C a are arbitrary constants. In a similar manner if we had assumed Y v = B v sin ut [see 
remarks at the end of Problem 8.30] we would have obtained 

avir 



>(«) _ 



D„ sin 



N+l 
Thus solutions are given by 

C a sin — cos u>t and D a sin — sin ut 

N+l N+l 

and since sums of solutions are also solutions, we have 

N 

*v — 2i sin \C a cos oxt + D a sin «t) 

a = l iV + 1 

The constants C a and D a are determined from initial conditions. 
The analogy with the continuous vibrating string is easily seen. 



Supplementary Problems 

VIBRATING SYSTEMS OF PARTICLES 

8.33. Find the normal frequencies of the vibrations in Problem 8.1, page 197, if the spring constants 
and masses are all different. 

8.34. Two equal masses m on a horizontal frictionless 
table as shown in Fig. 8-26 are connected by equal 
springs. The end of one spring is fixed at A 

and the masses are set into motion, (a) Set up the ^JL~^ m * m 

equations of motion of the system. (6) Find the A \ r Q'O'OQ^ Q ' QOOQ ^ Q 

normal frequencies of vibration, (c) Describe the 
normal modes of vibration. 

a». (» A = 3^± JI /, = 4±1 JI Fi * 8 - 26 

Air \ m ' z 4v y m 

8.35. Work Problem 8.34 if the spring constants and masses are different. 

8.36 Two equal masses m are attached to the ends of a spring of constant k which is on a horizontal 
frictionless table. If the masses are pulled apart and then released, prove that they will vibrate 
with respect to each other with period 2iryfm/2tc. 

8.37. Work Problem 8.36 if the masses are different and equal to M t and M 2 respectively. 

Ans. 2irV^/K where n = M 1 M 7 J{M l + M 2 ) 



218 



APPLICATIONS TO VIBRATING SYSTEMS, ROCKETS AND COLLISIONS [CHAP. 8 



8.38. In Fig. 8-27 equal masses m lying on a horizontal 
frictionless table are connected to each other and to 
fixed points A and B by means of elastic strings of 
constant tension T and length I. If the displacements 
from the equilibrium position AB of the masses are 
Y l and F 2 respectively, show that the equations of 
motion are given by 

Y, = k(Y 2 -2Y 1 ), Y 2 = k(Y 1 ~2Y 2 ) 

where k — ST I ml. 




Fig. 8-27 



.39. Prove that the natural frequencies of the vibration in Problem 8.38 are given 
respectively by 



JL /3T 
2w \ ml 

and describe the modes of vibration. 



and 



1 9T 



2tt V ml 



.40. Find the normal frequencies and normal modes of vibration for the system of 
particles of masses m x and m 2 connected by springs as indicated in Fig. 8-28. 



A 




CHANGING MASS. ROCKETS 

8.41. (a) Prove that the total distance traveled by the rocket of Problem 8.5 in time t is given by 



v <t + 



m 



at\ /m 
In 



V m o 



- -gt 2 
2 y 



(b) What is the maximum height which the rocket can reach and how long will it take to 
achieve this maximum height? 



8.42. Suppose that a rocket which starts from rest falls in a constant gravitational field. At the 
instant it starts to fall it ejects gas at the constant rate a in the direction of the gravitational 
field and at speed v with respect to the rocket. Find its speed after any time t. 

m ^ 



Ans. gt — v In 



at J 



8.43. How far does the rocket of Problem 8.42 travel in time tl 



Ans. — gt 2 



m — at 
t + ( — ) In 



m — at 
m 



8.44. Describe how Problem 8.42 can be useful in making a "soft landing" on a planet or satellite? 

8.45. Discuss the motion of a two-stage rocket, i.e. one in which one part falls off and the other 
rocket takes over. 

COLLISIONS OF PARTICLES 

8.46. A gun fires a bullet of mass m with horizontal velocity v into a block of wood of mass M which 
rests on a horizontal frictionless plane. If the bullet becomes embedded in the wood, (a) determine 
the subsequent velocity of the system and (b) find the loss in kinetic energy. 

Ans. (a) mv/(M + m) (b) mMv 2 /2(M + m) 

8.47. Work Problem 8.46 if the block is moving away from the gun with velocity V. 

8.48. A ball which is dropped from a height H onto a floor rebounds to a height h < H. Determine 
the coefficient of restitution. Ans. -Jh/H 



8.49. A mass m 1 traveling with speed v on a horizontal plane hits another mass m 2 which is at rest. 
If the coefficient of restitution is e , prove that there is a loss of kinetic energy equal to 
m 1 m 2 (l — e 2 )v 2 /2(m 1 + m 2 ). 



CHAP. 8] APPLICATIONS TO VIBRATING SYSTEMS, ROCKETS AND COLLISIONS 219 

8.50. A billiard ball strikes another billiard ball obliquely at an angle of 45° with their line of 
centers at the time of impact. If the coefficient of restitution is 1/2, find the angle at which 
the first ball will "bounce off". Ans. tan" 1 (3/5) 

8.51. Let the masses of two colliding particles be m 1 ,m 2 and their respective velocities before impact 
be v t ,v 2 . If the coefficient of restitution is e, prove that the loss in kinetic energy as a result 

of the collision is •= , (v x - v 2 ) 2 (l - c 2 ). 

8.52. Prove that the momentum which is transferred from the first particle of Problem 8.51 to the second 

is -_, (1 + e )(v 1 — v 2 ). 

8.53. A ball is dropped from a height h above a horizontal plane on to an inclined plane of angle a which 
is resting on the horizontal plane. Prove that if the coefficient of restitution is e, then the ball will 
next hit the incline at a point which is at a distance 4e(l + e)h sin « below the original point 
of impact. 



FOURIER SERIES, ODD AND EVEN FUNCTIONS, FOURIER SINE AND COSINE SERIES 

8.54. Graph each of the following functions and find their corresponding Fourier series using properties 
of even and odd functions wherever applicable. 

[ 8 < x < 2 
(a) f(x) = j_ 8 2 < x < 4 Period 4 (") fW ~ 4x ' ° < x < 10 > Period 10 

IX C-x -A^x^O (2x 0^x<3 

{b) f{X) = { x ^ x £ 4 Peri ° d 8 W m = | _ s<x<0 Period 6 

Ans . (o) W | (1- cos n,) s . n n^ _ 40 - 1 « 

(h\ 2 -— V C 1 ~ cos nir) r nvx ... 3 « \ 6(cos nv - 1) nirx 6 coanir . nirx\ 
{b) ** A n* C0S ~T id) 2 + „?> J *P COS "3 nV~ Sin -3-| 

8.55. In each part of Problem 8.54, tell where the discontinuities of f(x) are located and to what value 
the series converges at these discontinuities. 

Ans. (a) x = 0, ±2, ±4, ... ; (c) x = 0, ±10, ±20, . . . ; 20 

(b) no discontinuities (d) x — ±3, ±9, ±15, . . . ; 3 

(2- x < a; < 4 

8.56. Expand f(x) = < in a Fourier series of period 8. 

[1-6 4<«<8 

i 6 - J „ E®. j_ 1. <?£» J^ 5 ffa; 
3 2 cos 4 ■+■ 52 



Ans. JHcos^f + ^cos^F + ^- cos ^ + 

8.57. (a) Expand f(x) = cos x, < x < v, in a Fourier sine series. 

(6) How should f(x) be defined at x = and a; = v so that the series will converge to fix) for 

=5 £C ^ 7T? 

a 1 \ 8 -v ttsin2/i# / ,, ,,^ ,, ^ 

Ans. (a) - n 2 4n2-1 (6) /(0) = fW) = 

8.58. (a) Expand in a Fourier series f(x) - cos a;, < x < v if the period is w, and (6) compare with 
the result of Problem 8.57, explaining the similarities and differences if any. 

Ans. Answer is the same as in Problem 8.57. 



220 APPLICATIONS TO VIBRATING SYSTEMS, ROCKETS AND COLLISIONS [CHAP. 8 

fx < x < 4 

8.59. Expand f{x) — < n J in a series of (a) sines, (6) cosines. 

[8 — x 4 < x < 8 

. , . 32 £, 1 , «ir . nvx /L , 16 « / 2 cos n-ir/2 — cos mr — l\ mrx 
Ans. (a) v% 2 x ^ Bm T sin— (6) ^2 ( - t ) cos — 

THE VIBRATING STRING 

8.60. (a) Solve the boundary-value problem 

d 2 Y d 2 Y 

1W = 4 M 0<x<,,t>0 

Y x (0,t) = 0, Y(r,t) = h, Y(x,0) = 0, Y t (x,0) = 

(6) Give a physical interpretation of the problem in (a). 

Aws. F(a, *) = — 2 „ ' sin (w - 4)« sin (2n - l)t 

t „=i 2w - 1 2 

8.61. Solve the boundary-value problem 

Y tt = Y xx - g < x < *, t > 

F(0, t) = 0, Y{ir,t) - 0, Y(x,0) = px(ir-x), Y t (x,0) = 
and interpret physically. 

Ans. Y(x, t) = 4(2/ta? + flr) % /o 1 ,^ sin ( 2w ~ !)* cos ( 2w _ !)* ~ i^"' _ x ) 

ir n =i (2w — 1)^ 

d 2 Y d 2 Y 

8.62. (a) Find a solution of the equation -rp- = 4 -^j which satisfies the conditions F(0, t) = 0, 

Y( v , t) - 0, Y(x, 0) = 0.1 sin x + 0.01 sin 4x, Y t (x, 0) = for < x < ir, t > 0. (6) Interpret physi- 
cally the boundary conditions in (a) and the solution. 
Ans. (a) Y(x, t) = 0.1 sin x cos 2t + 0.01 sin 4a; cos St 

Q2V d 2 Y 

8.63. (a) Solve the boundary-value problem -^ = 9 -^ subject to the conditions Y(0, t) = 0, 
F(2, t) = 0, Y(x, 0) = 0.05x(2 - x), Y t (x, 0) = 0, where < x <2,t> 0. (6) Interpret physically. 

/ x v , *v 1-6 4 1 • (2n - l)irx 3(2n-lM 

Aws. (a) F(ar, t) = -^j 2 ( 2w _ i)3 sm <j C0S 2 

8.64. Solve Problem 8.63 with the boundary conditions for Y(x,0) and Y t (x,0) interchanged, i.e. 
Y(x, 0) = 0, Y t (x, 0) = 0.05sc(2 - x), and give a physical interpretation. 

™ * 3.2 2, 1 . (2w - l)yg . B(2n-l)wt 
Ans. Y(x, t) =—t 2 x (2^71)4 ««n 2 «* 2 

MISCELLANEOUS PROBLEMS 

8.65. A spherical raindrop falling in a constant gravitational field grows by absorption of moisture from 
its surroundings at a rate which is proportional to the instantaneous surface area. Assuming 
that it starts with radius zero, determine its acceleration. Ans. ^g 

8.66. A cannon of mass M rests on a horizontal plane having coefficient of friction ft. It fires a 
projectile of mass m with muzzle velocity v in a direction making angle a with the horizontal. 
Determine how far back the cannon will move due to the recoil. 

8.67. A ball is thrown with speed v onto a smooth horizontal plane in a direction making angle a with 
the plane. If e is the coefficient of restitution, prove that the velocity of the ball after the impact 
is given by v V 1 - (1 — e 2 ) sin 2 a in a direction making angle tan -1 (e tan a) with the horizontal. 



CHAP. 8] APPLICATIONS TO VIBRATING SYSTEMS, ROCKETS AND COLLISIONS 



221 



8.68. Prove that the total theoretical time taken for the particle of Problem 8.67 to come to rest is 

8.69. Prove that while the particle of Problem 8.27, page 213, moves from the top to the bottom of the 
incline, the incline moves a distance {mL cos a)/{M + m). 

8.70. Prove that the loss of kinetic energy of the spheres of Problem 8.9 is ^(^i cos 0y — v 2 cos 2 ) 2 (1 ~~ e *) 
where fi is the reduced mass m 1 m 2 /{m, 1 + m 2 ). 



8.71. Prove that the acceleration of a double incline of mass M [Fig. 8-29] which is on a smooth 
(m 1 sin a x cos ay — m 2 sin o 2 cos a 2 )g 



table is given by 



M + my sin 2 ay + m 2 sin 2 a 2 



8.72. If A is the acceleration of the incline in Problem 8.71, prove that the accelerations of the masses 

my(A cos ay + g sin ay) + m 2 (A cos a 2 — g sin a 2 ) 

relative to the incline are given numerically by ; . 

m[ + m 2 





Fig. 8-29 



Pig. 8-30 



8.73. A mass in slides down an inclined plane of the same mass which is on a horizontal plane with 
coefficient of friction fi. Prove that the inclined plane moves to the right with acceleration equal 
to (1 - 3/t)/(3 - /x). See Fig. 8-30. 



8.74. A gun of mass M is located on an incline of angle a which in turn is on a smooth horizontal plane. 
The gun fires a bullet of mass m horizontally away from the incline with speed v . Find the 
recoil speed of the gun. Ans. (mv cos a)/M up the incline 

8.75. How far up the plane will the gun of Problem 8.74 move before it comes to rest if the incline is 
(a) frictionless, (b) has coefficient of friction /*? 

8.76. A weight W is dropped from a height H above the plate AB of 
Fig. 8-31 which is supported by a spring of constant k. Find the ^ 
speed with which the weight rebounds. 

8.77. A ball is thrown with speed v at angle a with a horizontal plane. 
If it rebounds successively from the horizontal plane, determine 
its location after n bounces. Assume that the coefficient of restitution 
is e and that air resistance is negligible. 



Fig. 8-31 



8.78. Work Problem 8.77 if the horizontal plane is replaced by an inclined plane of angle /? and the 
ball is (a) thrown downward, (b) thrown upward. 

8.79. Obtain the equation (1), page 195, for the vibrating string by considering the equations of motion 
for the N particles of Problem 8.29, page 215, and letting N -» °°. 

8.80. Prove that as N -* °° the normal frequencies as given in Problem 8.31, page 216, approach those 
for the continuous vibrating string. 



8.81. Prove that for ^ x ^ ir, 

t 2 f cos 1x 



(a) x(ir — x) = 
(6) x(v — x) — 



6 



l 2 



cos 4a; cos 6a; 

i ~m "i" T^o T 



2 2 



32 



8 /sin x , sin 3a; sin 5a; 



l 3 



33 



5 3 



222 APPLICATIONS TO VIBRATING SYSTEMS, ROCKETS AND COLLISIONS [CHAP. 8 

8.82. Use Problem 8.81 to show that 

n =i w 6 ~ x n 2 12 n =i (2n — l) 3 32 

8.83. Prove that Y = f(x + ct) + g(x - ct) is a solution of the equation 

d*Y _ „ d 2 Y 
dt 2 ° dx 2 

and discuss the connection of this solution with the problem of the vibrating string. 

T C x /dY\ 2 

8.84. (a) Prove that the total potential energy of a vibrating string is V = — I ( -r— ) dx. 

(b) Thus show that V = ~- 2 n% ( a n cos ~^i •" ^n si n — T - ) • 



2 



8.85. (a) Prove that the total kinetic energy of the vibrating string is K.E. = ^<r I ( — ) 

(b) Thus show that K.E. = .. 2 n2 ( a n cos ~i b n sin n J . 

(c) Can the kinetic energy be infinite? Explain. 

ir 2 T °° 

8.86. Prove that the total energy of a vibrating string is E = — tt- 2 n2 ( a n + &!»)• 

4f n=l 

8.87. Find the potential energy, kinetic energy and total energy for the string of (a) Problem 8.20, 
page 210, (b) Problem 8.28, page 214. 

8.88. If damping proportional to the instantaneous transverse velocity is taken into account in the 

d 2 Y dY „ d 2 Y 

problem of the vibrating string, prove that its equation of motion is -ttj- + P ~qT — c z T-g" • 

8.89. Prove that the frequencies of vibration for the damped string of Problem 8.88 are given by 
y/nW&IP - p 2 /4, n = 1, 2, 3, 

8.90. Solve the problem of the damped vibrating string if the string is fixed at the endpoints x = and 
x = I and the string is (a) given an initial shape f(x) and then released, (b) in the equilibrium 
position and given an initial velocity distribution g(x), (c) given an initial shape f(x) and velocity 
distribution g{x). 

8.91. Work the problem of the damped vibrating string if gravitation is taken into account. 

8.92. Work (a) Problem 8.84(a), (6) Problem 8.85(a), (c) Problem 8.86, (a*) Problem 8.88 for the 
case where the string is replaced by N particles as in Problem 8.29, page 215. 

8.93. In Fig. 8-32 the double pendulum system is free to vibrate in a vertical Y/////////// 
plane. Find the normal frequencies and normal modes assuming small 
vibrations. 

8.94. Work Problem 8.93 if there is an additional mass m 3 suspended from m 2 by 
a string of length Z 3 . 

8.95. Generalize the motion of (a) Problem 8.1, (6) Problem 8.34 to N equal 
particles and springs. 

8.96. In Problem 8.95 investigate the limiting case as N -» ». Discuss the 
physical significance of the results. 



CHAP. 8] APPLICATIONS TO VIBRATING SYSTEMS, ROCKETS AND COLLISIONS 223 

8.97. Solve the boundary-value problem 

d 2 Y _ fl2 F 

F(o,t) = o, F(U) = o, r(*,o) = /(«), Y t (*,o) = o 

and give a physical interpretation. 

8.98. Work Problem 8.97 if the condition Y t {x, 0) = is replaced by Y t (as, 0) = g(x). 

8.99. Work Problem 8.97 if the partial differential equation is replaced by 

and interpret physically. 



d*Y , Q dY 2 d*Y . 

■W + Pin = c Jx^ + asmut 



8.100. Set up the differential equations and initial conditions for the motion of a rocket in an inverse 
square gravitational field. Do you believe these equations can be solved? Explain. 

8.101. Two bodies [such as the sun and earth or earth and moon] of masses m x and m 2 move relative 
to each other under their mutual inverse square attraction according to the universal law of 
gravitation. If r x and r 2 are their position vectors relative to a fixed coordinate system, and 
r = r t — r 2 , prove that their equations of motion are given by 

.. _ Gm 2 (r 1 - r 2 ) .. _ Gm 1 {r 2 - r t ) 

r i ~ ~ r3 ' r 2 ~ ^3 

This is called the problem of two bodies. 

8.102. In Problem 8.101 choose a new origin at the center of mass of the two bodies, i.e. such that 
m 1 r 1 + m 2 r 2 = 0. Thus show that if we let r be the position vector of m x relative to m 2 , then 

.. _ G(m x + m 2 )r x .. G^ + m 2 )r 2 

r i ~ ^3 ' r 2 ~ ^ 

,. .. .. G{m 1 + m 2 )r 

or, on subtracting, r = ; 

r 3 

Thus show that the motion of m x relative to m 2 is exactly the same as if the body of mass m 2 were 
fixed and its mass increased to m l + m 2 . 

8.103. Using Problem 8.102, obtain the orbit of mass m t relative to w 2 and compare with the results of 
Chapter 5. Are Kepler's first and second laws modified in any way? Explain. 

8.104. If P is the period of revolution of m l about m 2 and a is the semi-major axis of the elliptical path of 
m 1 about m 2 , prove that 

P2 _ 4tt 2 

a 3 G(m l + m 2 ) 

Compare this result with Kepler's third law: In the case of the earth [or other planet] and sun, 
does this modified Kepler law have much effect? Explain. 

8.105. Set up equations for describing the motion of 3 bodies under a mutual inverse square law of 
attraction. 

8.106. Transform the equations obtained in Problem 8.105 so that the positions of the bodies are described 
relative to their center of mass. Do you believe these equations can be solved exactly? 

8.107. Work Problems 8.105 and 8.106 for N bodies. 



Chapter 9 PLANE MO TION 

of RIGID BODIES 



RIGID BODIES 

A system of particles in which the distance between any two particles does not change 
regardless of the forces acting is called a rigid body. Since a rigid body is a special case 
of a system of particles, all theorems developed in Chapter 7 are also valid for rigid bodies. 



TRANSLATIONS AND ROTATIONS 

A displacement of a rigid body is a change from one position to another. If during 
a displacement all points of the body on some line remain fixed, the displacement is called 
a rotation about the line. If during a displacement all points of the rigid body move in 
lines parallel to each other the displacement is called a translation. 



EULER'S THEOREM. INSTANTANEOUS AXIS OF ROTATION 

The following theorem, called Euler's theorem, is fundamental in the motion of rigid 
bodies. 

Theorem 9.1. A rotation of a rigid body about a fixed point of the body is equivalent 
to a rotation about a line which passes through the point. 

The line referred to is called the instantaneous axis of rotation. 

Rotations can be considered as finite or infinitesimal. Finite rotations cannot be 
represented by vectors since the commutative law fails. However, infinitesimal rotations 
can be represented by vectors. 



GENERAL MOTION OF A RIGID BODY. CHASLE'S THEOREM 

In the general motion of a rigid body, no point of the body may be fixed. In such case 
the following theorem, called Chasle's theorem, is fundamental. 

Theorem 9.2. The general motion of a rigid body can be considered as a translation 
plus a rotation about a suitable point which is often taken to be the center of mass. 



PLANE MOTION OF A RIGID BODY 

The motion of a rigid body is simplified considerably when all points move parallel 
to a given fixed plane. In such case two types of motion, called plane motion, are possible. 

1. Rotation about a fixed axis. In this case the rigid body rotates about a fixed axis 
perpendicular to the fixed plane. The system has only one degree of freedom [see Chap- 
ter 7, page 165] and thus only one coordinate is required for describing the motion. 

224 



CHAP. 9] 



PLANE MOTION OF RIGID BODIES 



225 



2. General plane motion. In this case the motion can be considered as a translation 
parallel to the given fixed plane plus a rotation about a suitable axis perpendicular to 
the plane. This axis is often chosen so as to pass through the center of mass. The num- 
ber of degrees of freedom for such motion is 3: two coordinates being used to describe 
the translation and one to describe the rotation. 

The axis referred to is the instantaneous axis and the point where the instantaneous 
axis intersects the fixed plane is called the instantaneous center of rotation [see page 229]. 

We shall consider these two types of plane motion in this chapter. The motion of a 
rigid body in three dimensional space is more complicated and will be considered in 
Chapter 10. 



MOMENT OF INERTIA 

A geometric quantity which is of great importance in discussing the motion of rigid 
bodies is called the moment of inertia. 

The moment of inertia of a particle of mass m about a line or axis AB is defined as 



/ = mr 2 

where r is the distance from the mass to the line. 

The moment of inertia of a system of particles, with 
masses mi, ra 2 , . . . , m N about the line or axis AB is defined as 



/ = ^m v r 2 = m t r 2 + m 2 r 2 + 



+ m N r N 



(2) 



where r%, r 2 , . . . , r N are their respective distances from AB. 

The moment of inertia of a continuous distribution of 
mass, such as the solid rigid body % of Fig. 9-1, is given by 

/ = J r 2 dm (S) 

where r is the distance of the element of mass dm from AB. 




(-0 



Fig. 9-1 



RADIUS OF GYRATION 

N 

Let / = ^ m v r 2 be the moment of inertia of a system of particles about AB, and 

N v = i 

M = ^ m v be the total mass of the system. Then the quantity K such that 

2 ™«> r l 



K 2 = — = 
M 



2 m * 



(*) 



is called the radius of gyration of the system about AB. 
For continuous mass distributions (4) is replaced by 



K 2 = — = 



Jr 2 dm 

J dm 



(5) 



226 



PLANE MOTION OF RIGID BODIES 



[CHAP. 9 



THEOREMS ON MOMENTS OF INERTIA 

1. Theorem 9.3: Parallel Axis Theorem. Let / be the moment of inertia of a system 
about axis AB and let I c be the moment of inertia of the system about an axis parallel to 
AB and passing through the center of mass of the system. Then if b is the distance between 
the axes and M is the total mass of the system, we have 



/ = 



I c + Mb 2 



2. Theorem 9.4: Perpendicular Axes Theorem. 

xy plane of an xyz coordinate system. Let h, I y and h 
the x, y and z axes respectively. Then 

h = I X +Iy 



Consider a mass distribution in the 
denote the moments of inertia about 



en 



SPECIAL MOMENTS OF INERTIA 

The following table shows the moments of inertia of various rigid bodies which arise in 
practice. In all cases it is assumed that the body has uniform [i.e. constant] density. 



Rigid Body 


Moment of Inertia 


1. Solid Circular Cylinder 

of radius a and mass M 
about axis of cylinder. 


\Ma? 


2. Hollow Circular Cylinder 
of radius a and mass M 
about axis of cylinder. 
Wall thickness is negligible. 


Ma* 


3. Solid Sphere 

of radius a and mass M 
about a diameter. 


|Ma2 


4. Hollow Sphere 

of radius a and mass M 

about a diameter. 

Sphere thickness is negligible. 


Ma2 


5. Rectangular Plate 

of sides a and 6 and 
mass M about an axis 
perpendicular to the plate 
through the center of mass. 


JLM(a2 + 62) 


6. Thin Rod 

of length a and mass M 
about an axis perpendicular 
to the rod through the 
center of mass. 


^Ma* 



COUPLES 

A set of two equal and parallel forces which act in 
opposite directions but do not have the same line of action 
[see Fig. 9-2] is called a couple. Such a couple has a turning 
effect, and the moment or torque of the couple is given 
by r X F. 

The following theorem is important. 




-F 



Fig. 9-2 



CHAP. 9] 



PLANE MOTION OF RIGID BODIES 



227 



Theorem 9.5. Any system of forces which acts on a rigid body can be equivalently 
replaced by a single force which acts at some specified point together with a suitable couple. 



KINETIC ENERGY AND ANGULAR MOMENTUM ABOUT A FIXED AXIS 

Suppose a rigid body is rotating about a fixed axis with 
angular velocity <•> which has the direction of the axis AB 
[see Fig. 9-3]. Then the kinetic energy of rotation is 
given by 

T = i/o> 2 (8) 

where / is the moment of inertia of the rigid body about 
the axis. 



Similarly the angular momentum is given by 

O = 7o> (9) 




Fig. 9-3 



MOTION OF A RIGID BODY ABOUT A FIXED AXIS 

Two important methods for treating the motion of a rigid body about a fixed axis are 
given by the following theorems. 

Theorem 9.6: Principle of Angular Momentum. If A is the torque or the moment of 
all external forces about the axis and O = I<a is the angular momentum, then 

d 



dt 



(/•) = IS, = It 



(10) 



A = 
where a is the angular acceleration. 

Theorem 9.7: Principle of Conservation of Energy. If the forces acting on the rigid 
body are conservative so that the rigid body has a potential energy V, then 



T + V 



i/ w 2 + V = E = constant 



(«) 



WORK AND POWER 

Consider a rigid body % capable of rotating in a 
plane about an axis O perpendicular to the plane, as 
indicated in Fig. 9-4. If a is the magnitude of the torque 
applied to the body under the influence of force F at 
point A, the work done in rotating the body through 
angle de is 

dW = AdO (12) 

and the instantaneous power developed is 

dW 
dt 



cp - 



= A. 



(13) 




Fig. 9-4 



where w is the angular speed. 

We have the following 

Theorem 9.8. The total work done in rotating a rigid body from an angle 0i where 
the angular speed is Wl to angle 2 where the angular speed is « a is the difference in the 
kinetic energy of rotation at w x and o> 2 . In symbols, 



J Ade = £/<of - £/o>f 



(U) 



228 



PLANE MOTION OF RIGID BODIES 



[CHAP. 9 



IMPULSE. CONSERVATION OF ANGULAR MOMENTUM 

The time integral of the torque 

J»t 2 
Adt 
h 

is called the angular impulse from time U to t 2 . 
We have the following theorems. 



(15) 



Theorem 9.9. The angular impulse is equal to the change in angular momentum. In 
symbols tf 

I Adt 



O2 — Oi 



(16) 



Theorem 9.10: Conservation of Angular Momentum. If the net torque applied to a 
rigid body is zero, then the angular momentum is constant, i.e. is conserved. 



THE COMPOUND PENDULUM 

Let % [Fig. 9-5] be a rigid body which is free to oscillate 
in a vertical plane about a fixed horizontal axis through O 
under the influence of gravity. We call such a rigid body a 
compound pendulum. 

Let C be the center of mass and suppose that the angle 
between OC and the vertical OA is 0. Then if h is the 
moment of inertia of % about the horizontal axis through O, 
M is the mass of the rigid body and a is the distance OC, 
we have for the equation of motion, 



H f— sm0 

Jo 




= 



(17) 



Fig. 9-5 



For small oscillations the period of vibration is 

P = 2ir\/Io/Mga (18) 

The length of the equivalent simple pendulum is 

I = hi Ma (19) 

The following theorem is of interest. 

Theorem 9.11. The period of vibration of a compound pendulum is a minimum when 
the distance OC = a is equal to the radius of gyration of the body about the horizontal 
axis through the center of mass. 



GENERAL PLANE MOTION OF A RIGID BODY 

The general plane motion of a rigid body can be considered as a translation parallel to 
the plane plus a rotation about a suitable axis perpendicular to the plane. Two important 
methods for treating general plane motion of a rigid body are given by the following 
theorems. 

Theorem 9.12: Principle of Linear Momentum. If r is the position vector of the center 
of mass of a rigid body relative to an origin O, then 



jfi (Mr) = Mr = F 



(20) 



where M is the total mass, assumed constant, and F is the net external force acting on 
the body. 



CHAP. 9] 



PLANE MOTION OF RIGID BODIES 



229 



Theorem 9.13. Principle of Angular Momentum. If I c is the moment of inertia of the 
rigid body about the center of mass, o> is the angular velocity and A c is the torque or 
total moment of the external forces about the center of mass, then 



= Tt^ 



>) 



= 7 c" 



(21) 



Theorem 9.14. Principle of Conservation of Energy. If the external forces are conser- 
vative so that the potential energy of the rigid body is V, then 

T + V = %mr 2 + i/ c o 2 + V = E = constant (22) 

Note that imr 2 = \mv 2 is the kinetic energy of translation and i/ c *» 2 is the kinetic 
energy of rotation of the rigid body about the center of mass. 



INSTANTANEOUS CENTER. 
SPACE AND BODY CENTRODES 

Suppose a rigid body % moves parallel to a given 
fixed plane, say the xy plane of Fig. 9-6. Consider 
an x'y' plane parallel to the xy plane and rigidly 
attached to the body. 

As the body moves there will be at any time t 
a point of the moving x'y' plane which is instan- 
taneously at rest relative to the fixed xy plane. 
This point, which may or may not be in the body, 
is called the instantaneous center. The line perpen- 
dicular to the plane and passing through the instan- 
taneous center is called the instantaneous axis. 




O 



Fig. 9-6 



As the body moves, the instantaneous center also moves. The locus or path of the 
instantaneous center relative to the fixed plane is called the space locus or space centrode. 
The locus relative to the moving plane is called the body locus or body centrode. The motion 
of the rigid body can be described as a rolling of the body centrode on the space centrode. 

The instantaneous center can be thought of as that point about which there is rotation 
without translation. In a pure translation of a rigid body the instantaneous center is 
at infinity. 



STATICS OF A RIGID BODY 

The statics or equilibrium of a rigid body is the special case where there is no motion. 
The following theorem is fundamental. 

Theorem 9.15. A necessary and sufficient condition for a rigid body to be in equilibrium 
is that 

F = 0, A = (23) 

where F is the net external force acting on the body and A is the net external torque. 



PRINCIPLE OF VIRTUAL WORK AND D'ALEMBERT'S PRINCIPLE 

Since a rigid body is but a special case of a system of particles, the principle of virtual 
work and D'Alembert's principle [see page 171] apply to rigid bodies as well. 



230 



PLANE MOTION OF RIGID BODIES 



[CHAP. 9 



PRINCIPLE OF MINIMUM POTENTIAL ENERGY. STABILITY 

At a position of equilibrium the net external force is zero, so that if the forces are 
conservative and V is the potential energy, 

F = -vV = (24) 



or in components, 



?I = dV dV_ 

dx U ' dy U ' dz ~ ° 



(25) 



In such case V is either a minimum or it is not a minimum. If it is a minimum the 
equilibrium is said to be stable and a slight change of the configuration will restore the 
body to its original position. If it is not a minimum the body is said to be in unstable 
equilibrium and a slight change of the configuration will move the body away from its 
original position. We have the following theorem. 

Theorem 9.16. A necessary and sufficient condition for a rigid body to be in stable 
equilibrium is that its potential energy be a minimum. 



Solved Problems 



RIGID BODIES 

9.1. A rigid body in the form of a triangle ABC 
[Fig. 9-7] is moved in a plane to position 
DEF, i.e. the vertices A, B and C are car- 
ried to D, E and F respectively. Show that 
the motion can be considered as a transla- 
tion plus a rotation about a suitable point. 

Choose a point G on triangle ABC which cor- 
responds to the point H on triangle DEF. Perform 
the translation in the direction GH so that triangle 
ABC is carried to A'B'C. Using H as center of 
rotation perform the rotation of triangle A'B'C 
through the angle 9 as indicated so that A'B'C is 
carried to DEF. Thus the motion has been accom- 
plished by a translation plus a rotation. 




Fig. 9-7 



9.2. 



Give an example to show that finite rotations cannot be represented by vectors. 

Let A x represent a rotation of a body [such as the rectangular parallelepiped of Fig. 9-8(a)] 
about the x axis while A y represents a rotation about the y axis. We assume that such rotations 
take place in a positive or counterclockwise sense according to the right hand rule. 





R 

V 



tf 



r (6) 
Fig. 9-8 




CHAP. 9] 



PLANE MOTION OF RIGID BODIES 



231 






In Fig 9-8(a) we start with the parallelepiped in the indicated position and perform the rotation 
A x about the x axis as indicated in Fig. 9-8(6) and then the rotation about the » axis m indi- 
cated in Fig. 9-8(c). Thus Fig. 9-8(c) is the result of the rotation A x + A y on Fig. 9-8(o). 

In Fig 9-9(a) we start with the parallelepiped in the same position as in Fig. 9-8(a), but this 
time, we first perform the rotation A y about the y axis as indicated in Fig. 9-9(6) and then the 
rotation A x about the x axis as indicated in Fig. 9-9(c). Thus Fig. 9-9(c) is the result of the 
rotation A y + A x on Fig. 9-9(a). 

Since the position of the parallelepiped of Fig. 9-8(c) is not the same as that of Fig. 9-9(c), 
we conclude that the operation A x + A y is not the same as A y + A x . Thus the commutative 
law is not satisfied, so that A x and A y cannot possibly be represented by vectors. 



MOMENTS OF INERTIA 

9.3. Two particles of masses m x and ra 2 respectively are connected by a rigid massless 

rod of length a and move freely in a plane. Show that the moment of inertia of the 

system about an axis perpendicular to the plane and passing through the center of 

mass is iacl 2 where the reduced mass 

ix — m\m<2.l{m\ + W2). 

Let r x be the distance of mass mj from the 
center of mass C. Then a — r x is the distance 
of mass m 2 from C. Since C is the center 
of mass, 

m l r 1 = m 2 (a — r t ) from which r 




m^a 



mj + W2 



Fig. 9-10 



and a — r t 



w^a 



!»! + W^ 



Thus the moment of inertia about an axis through C is 



m x r\ + m 2 (a — r x ) 2 = m. 



m 2 a 



. + m 2 ( : / 

m x + m 2 / \ m i + m 2/ 



m x a \ 2 _ wii% 



fllj+W^ 



a 2 = ndP- 



9.4. Find the moment of inertia of a solid circular cylinder of radius a, height h and 
mass M about the axis of the cylinder. 

Method 1, using single integration. 

Subdivide the cylinder, a cross section of which appears 
in Fig. 9-11, into concentric rings one of which is the ele- 
ment shown shaded. The volume of this element is 
(Area) (thickness) = (2irrdr)(h) = 2rrrh dr 
and the element of mass is dm = 2varh dr. 
The moment of inertia of dm is 

r 2 dm — 2iror a h dr 
where a is the density, and thus the total moment of inertia is 




J id 



2war s h dr = ^waha A 



(1) 



Fig. 9-11 



232 



PLANE MOTION OF RIGID BODIES 



[CHAP. 9 



Then since the mass is 



we find / = |Ma 2 . 



M 






2worh dr 



aira 2 h 



Method 2, using double integration. 

Using polar coordinates (r, 8), we see from Fig. 9-12 that 
the moment of inertia of the element of mass dm distant 
r from the axis is 

r 2 dm = r 2 ahrdrdo = ahr* dr do 

since hr dr do is the volume element and a is the mass per 
unit volume (density). Then the total moment of inertia is 



J-»2ir s»a 
I ahr^drdo = \irohcfi 



(1) 



0=0 "r=0 

The mass of the cylinder is given by 




M 



s»2tt pa 
^6=0 ^r=0 



ahr dr do = aira 2 h 



Fig. 9-12 



which can also be found directly by noting that the volume of the cylinder is ira 2 h. Dividing 
equation (1) by (2), we find I/M = %a 2 or I = \Ma 2 . 



9.5. Find the radius of gyration, K, of the cylinder of Problem 9.4. 

Since K 2 = I/M = \a 2 , K = a/y/2 = \a^2. 



9-6. Find the (a) moment of inertia and (b) radius of gyration of a rectangular plate 
with sides a and b about a side. 

Method 1, using single integration. 

(a) The element of mass shaded in Fig. 9-13 is ab dx, and its moment of inertia about the y axis is 
(ab dx)x 2 = abx 2 dx. Thus the total moment of inertia is 



J ia 
abx 2 dx = %aba* 



Since the total mass of the plate is M = aba, we have I/M = -Ja 2 or / = $Ma 2 . 



.(b) K 2 = I/M = %a 2 or K = a/y[S = %ayfz. 

y 







dx — •» 










1 

b 

1 













































Fig. 9-13 



6 * 



dm = a dy dx 



Fig. 9-14 



Method 2, using double integration. 

Assume the plate has unit thickness. If dm = adydx is an element of mass [see Fig. 9-14], 
the moment of inertia of dm about the side which is chosen to be on the y axis is x 2 dm = ax 2 dy dx. 
Then the total moment of inertia is 

Xa s\h 
I ax 2 dydx = j^aba? 

The total mass of the plate is M = aba. Then, as in Method 1, we find / = \Ma 2 and K = $ayfs. 



CHAP. 91 



PLANE MOTION OF RIGID BODIES 



231 



9.7. Find the moment of inertia of a right circular cone of height h and radius a 
about its axis. 

Method 1, using single integration. 

The moment of inertia of the circular cylindrical 
disc one quarter of which is represented by PQR in 
Fig. 9-15 is, by Problem 9.4, 

\{irr 2 o dz)(r 2 ) = ±iror*dz 

since this disc has volume vr 2 dz and radius r. 



From Fig. 9-15, 



h — z 



r 

— or r 
a 



h a ™\ h 

Then the total moment of inertia about the z axis is 



h 






dz = JL- 



Also, 

M = 



dz — 



^jjTra 4 <rh 



±ira 2 ha 




Fig. 9-15 



Thus / = ^Ma 2 . 

Method 2, using triple integration. 

Subdivide the cone, one quarter of which is 
shown in Fig. 9-16, into elements of mass dm as indi- 
cated in the figure. 

In cylindrical coordinates (r, 8, z) the element of 
mass dm of the cylinder is dm — or dr de dz where 
a is the density. 

The moment of inertia of dm about the z axis is 

r 2 dm = or 3 dr de dz 



As in Method 1, 



h — z 



h 



ha \ a 

Then the total moment of inertia about the z axis is 

>h(a — r)/a 




Fig. 9-16 



The total mass of the cone is 



J»27r pa s»n,(a — r)/a, 
0=0 J r =0 ^z=0 

is 

*S0=O ^r=0 *^2=0 



dr de dz — ^iraAoh 



■Jiia—rj/r 



M = 



or dr de dz 



^■n-aPho 



which can be obtained directly by noting that the volume of the cone is \tra 2 h. 



Thus / 



^Ma 2 . 



9.8. Find the radius of gyration K of the cone of Problem 9.7. 

K 2 = I/M = ^a 2 and K = ay/^ = ^ay/SO. 



THEOREMS ON MOMENTS OF INERTIA 

9.9. Prove the parallel axis theorem [Theorem 9.3, page 226]. 

Let OQ be any axis and ACP a parallel axis through the centroid C and distant 6 from OQ. 
In Fig. 9-17 below, OQ has been chosen as the z axis so that AP is perpendicular to the xy plane 
at P. 



234 



PLANE MOTION OF RIGID BODIES 



[CHAP. 9 




If b x is a unit vector in the direction OP, 
then the vector OP is given by 

b = 6bi (1) 

where b is constant and is the distance between 
axes. 

Let r„ and x' v be the position vectors of 
mass m„ relative to O and C respectively. If r 
is the position vector of C relative to O then 
we have 

r v = x' v + f (jg) 

The total moment of inertia of all masses 
m v about axis OQ is 

N 

/ = 2 m^.bi)* (3) Fig. 9-17 

The total moment of inertia of all masses m v about axis ACP is 

N 

Ic = 2 m^-b^ 
Then using (2) we find 

N N 

I = 2 m^.bi) 2 = 2 mX'bi + fbi) 2 

K=l V=l 

N N N 

= 2 m^-b^ + 22 m^.b^Cf-bj) + 2 ^(f-b^ 

V=l V = l 1>=1 

/ N \ N 

= / c + 26 ( 2 m, < ) . b x + 6 2 2 m v = I c + Mb* 



N N 

since i'b l = b, ^ m v = M and 2 wh-rj = [Problem 7.16, page 178]. 

r=l v=l 

The result is easily extended to continuous mass systems by using integration in place of 
summation. 



(•4) 



9.10. Use the parallel axis theorem to find the moment of inertia 
of a solid circular cylinder about a line on the surface of 
the cylinder and parallel to the axis of the cylinder. 

Suppose the cross section of the cylinder is represented as in 
Fig. 9-18. Then the axis is represented by C, while the line on the 
surface of the cylinder is represented by A. 

If a is the radius of the cylinder, then by Problem 9.4 and the 
parallel axis theorem we have 

j A = I c + Ma?- = ±Ma 2 + Ma* - f Ma* 




Fig. 9-18 



9.11. Prove the perpendicular axes theorem [Theorem 
9.4, page 226]. 

Let the position vector of the particle with mass m v 
in the xy plane be 

r„ = x v i + yj 

[see Fig. 9-19]. The moment of inertia of m„ about the 
z axis is w^|r„| 2 . 

Then the total moment of inertia of all particles * 
about the z axis is 




Fig. 9-19 



CHAP. 9] 



PLANE MOTION OF RIGID BODIES 



235 



N N 

h =. 2«hr|r„|« = U(^ 2 + a 

v = \ v = l 

= 2 m v xl + 2 m,!^ = I~ + I„ 



v=l v =i 

where I x and J B are the total moments of inertia about the x axis and y axis respectively. 
The result is easily extended to continuous systems. 



9.12. Find the moment of inertia of a rectangular plate with sides a and b about an axis 
perpendicular to the plate and passing through a vertex. 

Choose the rectangular plate [see Fig. 9-20] in the 
xy plane with sides on the x and y axes. Choose the 
z axis perpendicular to the plate at a vertex. 

From Problem 9.6 we have for the moments of 
inertia about the x and y axes, 

I x = $Mb*, I y = ^Ma2 

Then by the perpendicular axes theorem the moment 
of inertia about the z axis is 



I z = I x + I y = £M(&2 + a 2) 
= $M(a 2 + b 2 ) 




Fig. 9-20 



COUPLES 

9.13. Prove that a force acting at a point of a rigid body can be equivalents replaced 
by a single force acting at some specified point together with a suitable couple. 

Let the force be F t acting at point P t as in 
Fig. 9-21. If Q is any specified point, it is seen that 
the effect of F 4 alone is the same if we apply two 
forces t t and — f x at Q. 

In particular if we choose f t = —F u i.e. if f 
has the same magnitude as Fj but is opposite in 
direction, we see that the effect of F x alone is the 
same as the effect of the couple formed by F t and 
f i - ~ F i [which has moment r,XF,] together with the 
force — fj = F 1# 




Fig. 9-21 



9.14. Prove Theorem 9.5 page 227: Any system of forces which acts on a rigid body can 
witlTsS cXr ^ 3 Single f0Ke WhiCH ^ at — «"*" "* « a er 



By Problem 9.13 we can replace the force F„ 
at P„ by the force P„ at Q plus a couple of moment 
r„ X F„. Then the system of forces P lf F 2 , . . . , F N at 
points P lt P 2 ,...,P N can be combined into forces 
Fj, F 2 , . . . , F N at Q having resultant 

F = F 1 + F 2 + • • • + Fn 

together with couples having moments 

'iXFj, r 2 XF 2 , ..., r N XF N 

which may be added to yield a single couple. Thus 
the system of forces can be equivalent^ replaced by 
the single force F acting at Q together with a couple 



.Py 




Fig. 9-22 



236 



PLANE MOTION OF RIGID BODIES 



[CHAP. 9 



KINETIC ENERGY AND ANGULAR MOMENTUM 

9.15. If a rigid body rotates about a fixed axis with 
angular velocity ©, prove that the kinetic energy 
of rotation is T = ^/w 2 where / is the moment of 
inertia about the axis. 

Choose the axis as AB in Fig. 9-23. A particle P of 
mass m v will rotate about the axis with angular speed «. 
Then it will describe a circle PQRSP with linear speed 
v v — ur v where r v is its distance from axis AB. Thus its 
kinetic energy of rotation about AB is ^m v v% = ±m v w 2 r 2 , 
and the total kinetic energy of all particles is 



T = 2 ^rn v ^rl 



-I ( 2 w„r 2 )<o 2 

v = l 



= ¥" 2 



where / = 2 w„ r 2 is the moment of inertia about AB. 

v = l 

The result could also be proved by using integration in 
place of summation. 




Fig. 9-23 



9.16. Prove that the angular momentum of the rigid body of Problem 9.15 is O = /». 

The angular momentum of particle P about axis AB is m v r*m. Then the total angular 
momentum of all particles about axis AB is 

N / N \ 

a - 2 m v r 2 , o> = ( 2 «iv r 2 ) » = /<* 



where I — 2 w„rf is the moment of inertia about AB. 

v = l 

The result could also be proved by using integration in place of summation. 



MOTION OF A RIGID BODY ABOUT A FIXED AXIS 

9.17. Prove the principle of angular momentum for a rigid body rotating about a fixed 

axis [Theorem 9.6, page 227]. 

By Problem 7.12, page 176, since a rigid body is a special case of a system of particles, 

A = dtl/dt where A is the torque or moment of all external forces about the axis and Q is the 

total angular momentum about the axis. 



Since O = /<■> by Problem 9.16, 



a _ d ir \ — T da 

A - di (Ia) ' ! Tt 



/i. 



9.18. Prove the principle of conservation of energy for a rigid body rotating about a 
fixed axis [Theorem 9.7, page 227] provided the forces acting are conservative. 

The principle of conservation of energy applies to any system of particles in which the forces 
acting are conservative. Hence in particular it applies to the special case of a rigid body rotating 
about a fixed axis. If T and V are the total kinetic energy and the potential energy, we thus have 



T + V = constant 



E 



Using the result of Problem 9.15, this can be written ^Iu> 2 + V - E. 



CHAP. 9] 



PLANE MOTION OF RIGID BODIES 



237 



WORK, POWER AND IMPULSE 

9.19. Prove equation (12), page 227, for the work done in rotating a rigid body about a 
fixed axis. 

Refer to Fig. 9-4, page 227. Let the angular velocity of the body be « = „k where k is a 
unit vector m the direction of the axis of rotation. The work done by F is 

fJr 

dW = F-tfr - F-f t dt = F-vd* = F.(.Xr)<ft 

= (rXF)-»(lf = A • m dt = A w dt = Ade 
where in the last two steps we use A = Ak, o> = «k and a - de/dt. 

9.20. Prove equation (18), page 227, for the power developed. 

From Problem 9.19 and the fact that de/dt = a , 

¥ = dW/dt = A de/dt = Aw 

9.21. Prove Theorem 9.8, page 227. 

deJl e dt h Zlhte Id ° /dt S ° tHat A = Idu/dt - Then from Problem 9 - 19 ^d the fact that 

ol Ade = \ l %» dt = l t '-*• = V4-M 

9.22. Prove Theorem 9.9, page 228: The angular impulse is equal to the change in 
angular momentum. 



f\dt = C- 

J tl J h dt 



dt = Q 2 - O x 



9.23. Prove Theorem 9.10, page 229, on the conservation of angular momentum if the net 
torque is zero. 

From Problem 9.22, if A = then Q 2 = Q v 



THE COMPOUND PENDULUM 

9.24. Obtain the equation of motion (17), page 228, for 
a compound pendulum. 
Method 1. 

Suppose that the vertical plane of vibration of the 
pendulum is chosen as the xy plane [Fig. 9-24] where 
the z axis through origin O is the horizontal axis of 
suspension. 

Let point C have the position vector a relative to O. 
Since the body is rigid, |a| = a is constant and is the 
distance from O to C. 

The only external force acting on the body is its 
weight Mg = -Mgj acting vertically downward. Thus 
we have 

A = total external torque about z axis 

= a X Mg = -a X Mgj = aMg sin e k (1) 
where k is a unit vector in the positive z direction [out of the plane of the paper toward the reader]. 
Also, the instantaneous angular velocity is 

_ i de . . 

« - -<ok = ~^k = -*k 




Fig. 9-24 



so that if I is the moment of inertia about the z axis 



238 PLANE MOTION OF RIGID BODIES [CHAP. 9 

O = angular momentum about z axis = /<><* — — -M k 
Substituting from (1) and (2) into A = dil/dt, 

aMg sin fk = |r(-7o*k) or IS + ^^ sin * = (5) 

Method 2. 

The force Mg = — Mgj is conservative, so that the potential energy V is such that 

_ vv = «E, _ *E j _ £ k = _»„ or £ = 0, f = *,, £ = 

da; fly dz 9» 92/ Sz 

from which V = Mgy + c = —Mga cos o + c (4) 

since 1/ = —a cos d. This could be seen directly since y = —a cos e is the height of C below the 
x axis taken as the reference level. 

By Problem 9.15, the kinetic energy of rotation is ±I u 2 - ^I <> 2 - Then the principle of 
conservation of energy gives 

T + V - |/ o 2 - Mga cos = constant = E (5) 

Differentiating equation (5) with respect to t, 

1 6 'S + Mga sin e — 

or, since is not identically zero, I e + Mga sin $ — as required. 



9.25. Show that for small vibrations the pendulum of Problem 9.24 has period 
P = 27T\/Mga/h. 

For small vibrations we can make the approximation sin = so that the equation of motion 
becomes „, 

%-+*¥*, = o a) 



Then, as in Problem 4.23, page 102, we find that the period is P = 2jry/I /Mga. 



9.26. Show that the length I of a simple pendulum equivalent to the compound pendulum 
of Problem 9.24 is I = I of Ma. 

The equation of motion corresponding to a simple pendulum of length I suspended vertically 
from O is [see Problem 4.23, equation (2), page 102] 

6 + j sin e - (-0 

Comparing this equation with (1) of Problem 9.25, we see that I = I /Ma. 



GENERAL PLANE MOTION OF A RIGID BODY 

9.27. Prove the principle of linear momentum, Theorem 9.12, page 228, for the general 
plane motion of a rigid body. 

This follows at once from the corresponding theorem for systems of particles [Theorem 7-1, 
page 167], since rigid bodies are special cases. 

9.28. Prove the principle of angular momentum, Theorem 9.13, page 229, for general 
plane motion of a rigid body. 

This follows at once from the corresponding theorem for systems of particles [Theorem 7-4, 
page 168], since rigid bodies are special cases. 



CHAP. 9] 



PLANE MOTION OF RIGID BODIES 



239 



9.29. A solid cylinder of radius a and 
mass M rolls without slipping down 
an inclined plane of angle a. Show 
that the acceleration is constant and 
equal to f g sin a. 

Suppose that initially the cylinder has 
point O in contact with the plane and that 
after time t the cylinder has rotated 
through angle [see Fig. 9-25]. 

The forces acting on the cylinder at 
time t are: (i) the weight Mg acting verti- 
cally downward at the center of mass C; 
(ii) the reaction R of the inclined plane act- 
ing perpendicular to the plane; (iii) the 
frictional force f acting upward along the 
incline. 




Fig. 9-25 



Choose the plane in which motion takes place as the xy plane, where the * axis is taken as 
positive down the incline and the origin is at O. 

If r is the position of the center of mass at time t, then by the principle of linear momentum, 



Mr = Mg + R + f 
But g = g sin a i - g cos a j, R = Rj, f = -/i. Hence (1) can be written 

Mr* = (Mg sin a - f)\ + (R - Mg cos «)j 
The total external torque about the horizontal axis through the center of mass is 
A = OXMg + OXR + CBXf = CBXf ^ (-aj)X(-/i) = - a /k 
The total angular momentum about the horizontal axis through the center of mass is 

O = / c o, = 7 c (-*k) = -7 c *k 
where I c is the moment of inertia of the cylinder about this axis. 

Substituting (3) and U) into A = dCl/dt, we find -a/k = -7 c Vk or I c e = af. 
Using r = xi + yj in (2), we obtain 

M x = Mg sin a ~ f, My = R - Mg cos a 



W 



(2) 



(3) 



U) 



(5) 



Nowjf there is no slipping, x = ae or = x/a. Similarly, since the cylinder remains on the 
incline, y = 0; hence from (5), R = Mg cos a. 

Using e = x/a in I c e = af, we have f = I c '£/a*. From Problem 9.4, I c = JLMa* Then 
substituting / = iMx into the first equation of (5), we obtain x = \g sin a as required. 

9.30. Prove that in Problem 9.29 the coefficient of friction must be at least * tan«. 

The coefficient of friction is n — f/R. 

From Problem 9.29 we have / = \M x = ±Mg sin a and B = Mg cos a. Thus in order that 
slipping will not occur, n must be at least f/R = 1 tan a. 



9.31. (a) Work Problem 9.29 if the coefficient of friction between the cylinder and inclined 
plane is ^ and (b) discuss the motion for different values of /*. 

(a) In equation (5) of Problem 9.29, substitute / = pR = ^Mg cos a and obtain 

*** = £f(sin a — ix cos a) 

Note that in this case the center of mass of the cylinder moves in the same manner as 
a particle sliding down an inclined plane. However, the cylinder may slip as well as roll. 



240 



PLANE MOTION OF RIGID BODIES 



[CHAP. 9 



The acceleration due to rolling is 



a 9 = 



a 2 f a 2 /iMg cos a 



= 2ng cos a. 



Ic \M& 

The acceleration due to slipping is x — ae = g(sin a — 3/* cos a). 

(b) If (sin a — 3/i cos a) > 0, i.e. u < ^ tan a, then slipping will occur. If (sin a — 3fi cos a) ^ 0, 
i.e. /* = ^ tan a, then rolling but no slipping will occur. These results are consistent with 
those of Problem 9.30. 

9.32. Prove the principle of conservation of energy [Theorem 9.14, page 229]. 

This follows from the corresponding theorem for systems of particles, Theorem 7-7, page 169. 
The total kinetic energy T is the sum of the kinetic energy of translation of the center of mass 
plus the kinetic energy of rotation about the center of mass, i.e., 

T = -imr 2 + |/ c <o 2 

If V is the potential energy, then the principle of conservation of energy states that if E 
is a constant, 

T + V = -|mf 2 + £/ c <o 2 + V = E 

9.33. Work Problem 9.29 by using the principle of conservation of energy. 

The potential energy is composed of the potential energy due to the external forces [in this 
case gravity] and the potential energy due to internal forces [which is a constant and can be 
omitted]. Taking the reference level as the base of the plane and assuming that the height of 
the center of mass above this plane initially and at any time t to be H and h respectively, we have 

IMf 2 + l/ c <o 2 + Mgh - MgH 



or, using H — h = x sin a 



and r 2 = x 2 + y 2 = x 2 since y = 0, 
\Mx 2 + ^7 c w 2 = Mgx sin a 



Substituting w = — x/a and I c — \Ma 2 , we find x 2 = \gx sin a. Differentiating with respect to 
t, we obtain 

2x x = §gx sin a or x = f g sin a 



INSTANTANEOUS CENTER. SPACE AND BODY CENTRODES 

9.34. Find the position vector of the instantaneous center for a rigid body moving 
parallel to a given fixed plane. 

Choose the XY plane of Fig. 9-26 as the fixed 
plane and the xy plane as the plane attached to 
and moving with the rigid body %. Let point P of 
the xy plane [which may or may not be in the 
rigid body] have position vectors R and r relative 
to the XY and xy planes respectively. If v and 
v^ are the respective velocities of P and A relative 
to the XY system, 

v = v A + »Xr = V A + o» X (R — R A ) (1) 

where R A is the position vector of A relative to O. 
If P is to be the instantaneous center, then v = 

so that 

• X(R-R A ) = -v A («) 

Multiplying both sides of (2) by o> X and using (7), Fig. 9-26 

page 5, 

«{«»(R — R A )} — (R — R A )(«»«) = -«Xv A 

Then since » is perpendicular to R — R A , this becomes 

(R - R A )w 2 = » X v A or R = R A + 




« X v A 



(3) 



CHAP. 9] 



PLANE MOTION OF RIGID BODIES 



241 



9.35. A cylinder moves along a horizontal plane. Find the (a) space centrode, (o) body 
centrode. Discuss the case where slipping may occur. 



(a) 



The general motion is one where both rolling 
and slipping may occur. Suppose the cylinder 
is moving to the right with velocity v A [the 
velocity of its center of mass] and is rotating 
about A with angular velocity «. 

Since o> = — wk and v A = v A i, we have 
• x v a = -wv A j so that (3) of Problem 9.34 
becomes . . . 

_ _ (« v a)j „ v A# 

R = R A - — X~ = R A 3 

In component form, 




Fig. 9-27 



Xi + Yj - X A i + oj - (v A /«)j x>r X = X A , Y = a - v A /a 

Thus the instantaneous center is located vertically above the point of contact of the cylinder 
with the ground and at height a — v A /u above it. 

Then the space centrode is a line parallel to the horizontal and at distance a — v A /u 
above it. If there is no slipping, then v A = aa and the space centrode is the X axis while 
the instantaneous center is the point of contact of the cylinder with the X axis. 

(b) The body centrode is given by |r| = v /o, or a circle of radius v /u. In case of no slipping, 
v = ao3 and the body centrode is the circumference of the cylinder. 



9.36. Solve Problem 9.29 by using the instantaneous center. 

By Problem 9.35, if there is no slipping then 
the point of contact P of the cylinder with the 
plane is the instantaneous center. The motion of 
P is parallel to the motion of the center of mass, 
so that we can use the result of Problem 7.86(c), 
page 191. 

The moment of inertia of the cylinder about 
P is, by the parallel axis theorem, ^Ma 2 + Ma 2 = 
2 Ma 2 . The torque about the horizontal axis through 
P is Mga sin a. Thus 

-^{lMa 2 'e) = Mga sine 

or e = -r^smfl 

6a 

Since x = ae, the acceleration is x = f g sin 9. 




Fig. 9-28 



STATICS OF A RIGID BODY 

9.37. A ladder of length I and weight Wi has one end 
against a vertical wall which is frictionless and 
the other end on the ground assumed horizontal. 
The ladder makes an angle a with the ground. 
Prove that a man of weight W m will be able to 
climb the ladder without having it slip if the 
coefficient of friction /* between the ladder and 



the ground is at least 



Wm + Wl 
Wm+Wr 



COt a. 



Let the ladder be represented by AB in Fig. 9-29 and 
choose an xy coordinate system as indicated. 




Fig. 9-29 



242 



PLANE MOTION OF RIGID BODIES 



[CHAP. 9 



The most dangerous situation in which the ladder would slip occurs when the man is at the 
top of the ladder. Hence we would require that the ladder be in equilibrium in such case. 

The forces acting on the ladder are: (i) the reaction R x = R t i of the wall; (ii) the weight 
W m = — W m j of the man; (iii) the weight W £ = — W t j of the ladder concentrated at C, the center of 
gravity; (iv) the reaction R 2 = R 2 j of tn © ground; (v) the friction force f = — /i. 

For equilibrium we require that 

F = 0, A = (I) 

where F is the total external force on the ladder and A the total external torque taken about a 
suitable axis which we shall take as the horizontal axis through A perpendicular to the xy plane. 
We have 

F = R x + W m + W, + R 2 + f = (R 1 -f)i + (-W m -W t + R 2 )j = 

if JBj — / = and -W m - W, + R 2 = (2) 

Also, A = (0) X Bi + (0) X W m + (AC) XW,+ (AB) X B 2 + (AB) X f 

= (0) X (flji) + (0) X (- W m j) + ($1 cos a i - ±1 sin a j) x (-TFJ) 

+ (I cos a i — I sin a j) X (R 2 j) + (I cos a i — I sin a j) X (— /i) 

= —tyWi cos a k + IR 2 cos a. k — If sin a k = 
—^Wi COS a + R 2 COS a — f sin a = 



if 



Solving simultaneously equations (2) and (3), we find 



/ = R l = (W TO + W l )cot« 



and 



R, = W m + W, 



Then the minimum coefficient of friction necessary to prevent slipping of the ladder is 

cot a 



/ _ w m + jW t 



Ra 



w m + w l 



(3) 



MISCELLANEOUS PROBLEMS 

9.38. Two masses mi and m 2 are connected by an inextensible string of negligible mass 
which passes over a frictionless pulley of mass M, radius a and radius of gyration K 
which can rotate about a horizontal axis through C perpendicular to the pulley. 
Discuss the motion. 

Choose unit vectors i and j in the plane of rotation 
as shown in Fig. 9-30. 

If we represent the acceleration of mass m x by Aj, 
then the acceleration of mass ra 2 is — Aj. 

Choose the tensions T x and T 2 in the string as shown 
in the figure. By Newton's second law, 

CO 
(2) 

(3) 
U) 



mg 



w-iAj = T x + m x g = — TJ + m^j 
-mgAj = T 2 + m 2 g = -T 2 j + m^j 

Thus m x A = mxQ — T lt m 2 A — T 2 — m 2 g 
or T t = m^g-A), T 2 = m 2 {g + A) 

The net external torque about the axis through C is 

A = (-oi) X (-TJ) + (ai) X (-r^) = a(7\ - T 2 )k 

The total angular momentum about O is 

CI = I c » = 7 c cok = IcOk 

Since A = dil/dt, we find from (5) and (6), 

a{T x -T 2 ) = I c 'o = MK*$ 




Fig. 9-30 



(5) 
(6) 
(7) 



CHAP. 9] 



PLANE MOTION OF RIGID BODIES 



243 



If there is no slipping about the pulley, we also have 

A — a'e 



Using (8) in (7), 
Using (4) in (9), 



Tx ~ T 2 



A = 



MK 2 



(wn - m 2 )g 



m 1 + m 2 + MkVa? 



(8) 
(9) 

(10) 



Thus the masses move with constant acceleration given in magnitude by (10). Note that if M — 0, 
the result (10) reduces to that of Problem 3.22, page 76. 



9.39. Find the moment of inertia of a solid sphere about a diameter. 

Let O be the center of the sphere and AOB be the 
diameter about which the moment of inertia is taken 
[Fig. 9-31]. Divide the sphere into discs such as QRSTQ 
perpendicular to AOB and having center on AOB at P. 

Take the radius of the sphere equal to a, OP = z, 
SP = r and the thickness of the disc equal to dz. Then 
by Problem 9.4 the moment of inertia of the disc about 
AOB is 

\(trr 2 o dz)r 2 - ^var^dz (1) 



From triangle OSP, r 2 = a 2 - z 2 . 
the total moment of inertia is 



Substituting into (1), 



= f 



\ira(a 2 - z 2 ) 2 dz 




The mass of the sphere is 

M 



-f_ 



•gwa 3 a 



7ra(a 2 - z 2 ) dz = 
2= —a 
which could also be seen by noting that the volume of the sphere is ova 3 . 



($) 



From (2) and (S) we have I/M = \a 2 ox I - %Ma 2 . 



9.40. A cube of edge s and mass M is suspended vertically from one of its edges, (a) Show 

that the period for small vibrations is P = 2v -^2y/s/Sg. (b) What is the length of 
the equivalent simple pendulum? 



(a) 



Since the diagonal of a square of side 8 has length V» 2 + s 2 = 8^/2, 
the distance OC from axis O to the center of mass is ^8^/2. 

The moment of inertia / of a cube about an edge is the same as 
that of a square plate about a side. Thus by Problem 9.6, 

/ = %M(s 2 + s 2 ) = §Ms 2 . 

Then the period for small vibrations is, by Problem 9.25, 

P = 2v *sl%Ms 2 /[Mg($ 8 V2)] = 2wyf2y[slZg~ 



(b) The length of the equivalent simple pendulum is, by Problem 9.26, 

I = %MsV[M(%sV2)] = fv^s 




Fig. 9-32 



9.41. Prove Theorem 9.11, page 228: The period of small vibrations of a compound 
pendulum is a minimum when the distance OC = a is equal to the radius of gyration 
of the body about a horizontal axis through the center of mass. 

If I c is the moment of inertia about the center of mass axis and I is the moment of inertia 
about the axis of suspension, then by the parallel axis theorem we have 



244 



PLANE MOTION OF RIGID BODIES 



[CHAP. 9 



7 = i c + Ma 2 
Then the square of the period for small vibrations is given by 



p2 = 



4fl- 2 /p _ 4ir 2 / h 



+ a 



2/Kr 



+ a 



Mga g \Ma J g 

where K* = I c / M is the square of the radius of gyration about the center of mass axis. 
Setting the derivative of P 2 with respect to a equal to zero, we find 

da^ ' g \ a? 

from which a - K c . This can be shown to give the minimum value since d 2 (P 2 )/da 2 < 0. Thus 
the theorem is proved. 

The theorem is also true even if the vibrations are not assumed small. See Problem 9.147. 



= 



9.42. A sphere of radius a and mass m rests on top of a fixed rough sphere of radius b. 
The first sphere is slightly displaced so that it rolls without slipping down the second 
sphere. Where will the first sphere leave the second sphere? 

Let the xy plane be chosen so as to pass 
through the centers of the two spheres, with the 
center of the fixed sphere as origin O [see Fig. 
9-33]. Let the position of the center of mass C of 
the first sphere be measured by angle 0, and sup- 
pose that the position vector of this center of 
mass C with respect to O is r. Let r l and 9 X be 
unit vectors as indicated in Fig. 9-33. 

Resolving the weight W = — mgj into com- 
ponents in directions r t and 9 lt we have [compare 
Problem 1.43, page 24] 

W = (WT^ + (W •*!)#! 

= (-mgj-rj)^ + (-mgj • 9 t ) 9 X 
= —mg sin o r t — mg cos e 9 t 

The reaction force N and frictional force f 
are N = Nr lt f = f9 v Using Theorem 9.12, 
page 228, together with the result of Problem 1.49, 
page 26, we have 

F = ma = m[(r — r'o 2 )r x + (r'e + 2ro)9 1 ] 

= W + N + £ 

= (N — mg sin 0)^ + (/ — mg cos 

re*) - N — mg sine, m(r'o + 2re) = f 




Fig. 9-33 



from which m(r 

Since r = a + b [the distance of C from O], these equations become 



= / 



mg cos e 



mg cos 6 



(1) 



—m(a + b)'e 2 = N — mg sin e, m(a + b) 

We now apply Theorem 9.13, page 229. The total external torque of all forces about the 
center of mass C is [since W and N pass through C], 

A = (-ar,) X f = (-ar 1 )X(f9 1 ) = -a/k 

Also, the angular acceleration of the first sphere about C is 

d 2 

a = -^(0 + v^)k = -(<£ + vOk 

Since there is only rolling and no slipping it follows that arc AP equals arc BP, or b$ = af. 
Then </> = v/2 - e and ^ = (b/a)(ir/2 — o), so that 



= -(0 + vOk 



-{ -6 -~ $ 

a 



^±-* )*k 



CHAP. 9] PLANE MOTION OF RIGID BODIES 245 

Since the moment of inertia of the first sphere about the horizontal axis of rotation through C 
is I — &ma 2 , we have by Theorem 9.13, 

A = la, -afk = |wa2(^-i-^ )Vk or / = -im(a+b)e 



Using this value of / in the second equation of (1), we find 

5g 

6 = "7(^+6) C0S6 W 

Multiplying both sides by $ and integrating, we find after using the fact that e = at t = or 
e = r/2, 

Using (*) in the first of equations (J), we find A/" = lwflr(17 sin - 10). Then the first sphere 
leaves the second sphere where N = 0, i.e. where 6 — sin-* 10/17. 




Supplementary Problems 

RIGID BODIES 

9.43. Show that the motion of region % of Fig. 9-34 can 
be carried into region %' by means of a translation 
plus a rotation about a suitable point. 

9.44. Work Problem 9.1, page 230, by first applying a 
translation of the point A of triangle ABC. 

9.45. If A X ,A V ,A Z represent rotations of a rigid body about 
the x, y and z axes respectively, is it true that the 
associative law applies, i.e. is A x + (A y + A z ) = 
(A x + A y ) + A z t Justify your answer. Fig. 9-34 

MOMENTS OF INERTIA 

9.46. Three particles of masses 3, 5 and 2 are located at the points (-1, 0, 1), (2, -1, 3) and (-2, 2, 1) 
respectively. Find (a) the moment of inertia and (6) the radius of gyration about the x axis. 
Ans. 71 

9.47. Find the moment of inertia of the system of particles in Problem 9.46 about (a) the y axis, 
(fe) the z axis. Ans. (a) 81, (6) 44 

9.48. Find the moment of inertia of a uniform rod of length I about an axis perpendicular to it and 
passing through (a) the center of mass, (b) an end, (c) a point at distance 1/4 from an end. 
Ans. (a) r&MP, (b) j^MP, (c) ^MP 

9.49. Find the (a) moment of inertia and (b) radius of gyration of a square of side a about a diagonal. 
Ans. (a) ^Ma*, (b) ^ay/S 

9.50. Find the moment of inertia of a cube of edge a about an edge. Ans. § Ma 2 

9.51. Find the moment of inertia of a rectangular plate of sides a and 6 about a diagonal. 

Ans. ±Ma 2 b 2 /(a 2 + b 2 ) 



9.52. 



Find the moment of inertia of a uniform parallelogram of sides a and 6 and included angle a 
about an axis perpendicular to it and passing through its center. Ans. J*M(a 2 + b 2 ) sin 2 a 



246 



PLANE MOTION OF RIGID BODIES 



[CHAP. 9 



9.53. 
9.54. 

9.55. 

9.56. 
9.57. 

9.58. 
9.59. 



Find the moment of inertia of a cube of side a about a diagonal. 

Find the moment of inertia of a cylinder of radius a and height h 
about an axis parallel to the axis of the cylinder and distant b from 
its center. Ans. ^M(a 2 + 2b 2 ) 

A solid of constant density is formed from a cylinder of radius a 
and height h and a hemisphere of radius a as shown in Fig. 9-35. 
Find its moment of inertia about a vertical axis through their 
centers. Ans. M(2a 3 + 15a 2 h)/{10a + 15h) 

Work Problem 9.55 if the cylinder is replaced by a cone of radius a 
and height h. 

Find the moment of inertia of the uniform solid region bounded 
by the paraboloid cz = x 2 + y 2 and the plane z = h about the 
z axis. Ans. ^Mch 




Fig. 9-35 



How might you define the moment of inertia of a solid about (a) a point, (6) a plane? Is there 
any physical significance to these results? Explain. 

Use your definitions in Problem 9.58 to find the moment of inertia of a cube of side a about 
(a) a vertex and (6) a face. Ans. (a) Ma 2 , (b) |ilfa 2 



KINETIC ENERGY AND ANGULAR MOMENTUM 

9.60. A uniform rod of length 2 ft and mass 6 lb rotates with angular speed 10 radians per second 
about an axis perpendicular to it and passing through its center. Find the kinetic energy of 
rotation. Ans. 100 lb ft 2 /sec 2 

9.61. Work Problem 9.60 if the axis of rotation is perpendicular to the rod and passes through an end. 
Ans. 400 lb ft 2 /sec 2 



9.62. 

9.63. 
9.64. 

9.65. 
9.66. 
9.67. 



A hollow cylindrical disk of radius a and mass M rolls along a horizontal plane with speed v. 
Find the total kinetic energy. Ans. Mv 2 



Work Problem 9.62 for a solid cylindrical disk of radius a. 



Ans. &Mv 2 



A flywheel having radius of gyration 2 meters and mass 10 kilograms rotates at angular 
speed of 5 radians/sec about an axis perpendicular to it through its center. Find the kinetic 
energy of rotation. Ans. 1000 joules 

Find the angular momentum of (a) the rod of Problem 9.60 (6) the flywheel of Problem 9.64. 
Ans. (a) 5 lb ft 2 /sec, (b) 200 kg m 2 /sec 

Prove the result of (a) Problem 9.15, page 236, (6) Problem 9.16, page 236, by using integration 
in place of summation. 

Derive a "parallel axis theorem" for (a) kinetic energy and (6) angular momentum and explain 
the physical significance. 



MOTION OF A RIGID BODY. THE COMPOUND PENDULUM. 
WORK, POWER AND IMPULSE 

9.68. A constant force of magnitude F is applied tangentially to a flywheel which can rotate about a 
fixed axis perpendicular to it and passing through its center. If the flywheel has radius a, radius 
of gyration K and mass M, prove that the angular acceleration is given by F a/MK 2 . 

9.69. How long will it be before the flywheel of Problem 9.68 reaches an angular speed w if it starts 
from rest? Ans. MK 2 a /F a 

9.70. Assuming that the flywheel of Problem 9.68 starts from rest, find (a) the total work done, 
(6) the total power developed and (c) the total impulse applied in getting the angular speed 
up to <o . Ans - ( a ) %MK 2 <»%> ( b ) *>«o. (c) MK 2 u 



CHAP. 9] 



PLANE MOTION OF RIGID BODIES 



247 



9.71. Work (a) Problem 9.68, (b) Problem 9.69 and (c) Problem 9.70 if F = 10 newtons, a = 1 meter, 
K - 0.5 meter, M = 20 kilograms and w = 20 radians/sec. 

Ans. (a) 2 rad/sec 2 ; (b) 10 sec; (c) 250 joules, 200 joules/sec, 100 newton sec 

9.72. Find the period of small vibrations for a simple pendulum assuming that the string supporting 
the bob is replaced by a uniform rod of length I and mass M while the bob has mass m. 



9.73. 
9.74. 



Ans. 2v a\ 



2(M + 3m)Z 



3(M + 2m)g 
Discuss the cases (a) M = and (b) m = in Problem 9.72. 

A rectangular plate having edges of lengths a and b respectively hangs vertically from the edge 
of length a. (a) Find the pe riod fo r small oscillations and (6) the length of the equivalent simple 
pendulum. Ans. (a) 2wy/2b/Zg , (b) |6 

9.75. A uniform solid sphere of radius a and mass M is suspended vertically downward from a point 
on its surface, (a) Find the period for small oscillations in a plane and (6) the length of the 
equivalent simple pendulum. Ans. (a) 2ir^la/hg, (b) 7a/5 



9.76. 



9.77. 



A yo-yo consists of a cylinder of mass 80 gm around which a string of length 60 cm is wound. 
If the end of the string is kept fixed and the yo-yo is allowed to fall vertically starting from rest, 
find its speed when it reaches the end of the string. Ans. 280 cm/sec 



Find the tension in the string of Problem 9.76. 
Ans. 19,600 dynes 



9.78. A hollow cylindrical disk of mass M moving with constant speed v 
comes to an incline of angle a. Prove that if there is no slipping 
it will rise a distance v\/{g sin a) up the incline. 

9.79. If the hollow disk of Problem 9.78 is replaced by a solid disk, how 
high will it rise up the incline? Ans. Svl/(4g sin a) 

9.80. In Fig. 9-36 the pulley, assumed frictionless, has radius 0.2 meter 
and its radius of gyration is 0.1 meter. What is the acceleration of 
the 5 kg mass? Ans. 2.45 m/sec 2 



Y/////////////////////' 



20 kg 



10 kg | | 



5 kg 



Fig. 9-36 



INSTANTANEOUS CENTER. SPACE AND BODY CENTRODES 

9.81. A ladder of length I moves so that one end is on a vertical wall and the other on a horizontal 
floor. Find (a) the space centrode and (b) the body centrode. 

Ans. (a) A circle having radius I and center at point O where the floor and wall meet. 
(b) A circle with the ladder as diameter 



9.82. 



9.83. 



9.84. 



A long rod AB moves so that it remains in contact with 
the top of a post of height h while its foot B moves on a 
horizontal line CD [Fig. 9-37]. Assuming the motion 
to be in one plane, find the locus of instantaneous centers. 

What is the (a) body centrode and (b) space centrode 
in Problem 9.82? 

Work Problems 9.82 and 9.83 if the post is replaced 
by a fixed cylinder of radius a. 




Fig. 9-37 



STATICS OF A RIGID BODY 

9.85. A uniform ladder of weight W and length I has its top against a smooth wall and its foot on a 
floor having coefficient of friction p. (a) Find the smallest angle a which the ladder can make 
with the horizontal and still be in equilibrium. (6) Can equilibrium occur if ft - 0? Explain. 



248 



PLANE MOTION OF RIGID BODIES 



[CHAP. 9 



9.86. Work Problem 9.85 if the wall has coefficient of friction n v 

9.87. In Fig. 9-38, AB is a uniform bar of length I and weight W supported at C. It carries weights W x 
at A and W 2 at D so that AC — a and CD — b. Where must a weight W 3 be placed on AC 
so that the system will be in equilibrium? 






w 3 
□ c 



A 

Fig. 9-38 



W 2 



D 






Fig. 9-39 



9.88. A uniform triangular thin plate hangs from a fixed point O by strings OA, OB and OC of 
lengths a, b and c respectively. Prove that the tensions T it T % and T 3 in the strings are such that 
T x /a = T 2 /b = T 3 /c. 

9.89. A uniform plank AB of length I and weight W is supported 
at points C and D distant a from A and b from B respec- 
tively [Fig. 9-39]. Determine the reaction forces at C and D 
respectively. 

9.90. In Fig. 9-40, OA and OB are uniform rods having the same 
density and connected at O so that AOB is a right angle. 
The system is supported at O so that AOB is in a vertical 
plane. Find the angles a and /? for which equilibrium occurs. 
Ans. a = tan -1 (a/b), /? = v/2 — tan -1 (a/b) 




Fig. 9-40 



MISCELLANEOUS PROBLEMS 

9.91. A circular cylinder has radius a and height h. Prove that the moment of inertia about an axis 
perpendicular to the axis of the cylinder and passing through the centroid is -J^M(h 2 + 3a 2 ). 

9.92. Prove that the effect of a force on a rigid body is not changed by shifting the force along its line 
of action. 

9.93. A cylinder of radius a and radius of gyration K rolls without slipping down an inclined plane of 
angle a and length I, starting from res t at the top of the in cline. Prove that when it reaches the 
bottom of the incline is speed will be ^(2gla 2 sin a)l(a 2 + K 2 ) . 

9.94. A cylinder resting on top of a fixed cylinder is given a slight displacement so that it rolls without 
slipping. Determine where it leaves the fixed cylinder. 

Ans. — sin -1 4/7 where e is measured as in Fig. 9-33, page 244. 



9.95. Work Problem 9.42 if the sphere is given an initial speed v . 

9.96. Work Problem 9.94 if the cylinder is given an initial speed v . 

9.97. A sphere of radius a and radius of gyration K about a diameter rolls without slipping down 
an incline of angle a. Prove that it descends with constant acceleration given by (ga 2 sin a)/(a 2 + K 2 ). 

9.98. Work Problem 9.97 if the sphere is (a) solid, (b) hollow and of negligible thickness. 
Ans. (a) f g sin a, (&) f g sin a 

9.99. A hollow sphere has inner radius a and outer radius 6. Prove that if M is its mass, then the 
moment of inertia about an axis through its center is 

'a 4 + a 3 & + a 2 & 2 + a& 3 + & 4> 



|M 



i 2 + ab + &2 



Discuss the cases 6 = and a — b. 



CHAP. 9] 



PLANE MOTION OF RIGID BODIES 



249 



9.100. Wooden plates, all having the same rectangular 
shape are stacked one above the other as indicated 
in Fig. 9-41. (a) If the length of each plate is 
2a, prove that equilibrium conditions will prevail 
if the (n + l)th plate extends a maximum distance 
of a/n beyond the wth plate where n = 1,2,3, ... . 
(b) What is the maximum horizontal distance 
which can be reached if more and more plates 
are added? 

9.101. Work Problem 9.100 if the plates are stacked on 
a sphere of radius R instead of on a flat surface 
as assumed in that problem. 



9.107. 



Fig. 9-41 



9.102. A cylinder of radius a rolls on the inner surface of a smooth cylinder of radius 2a. Prove that 
the period of small oscillations is 2v 1 \/3a/2g . 

9.103. A ladder of length I and negligible weight rests with one end against a wall having coefficient of 
friction ^ and the other end against a floor having coefficient of friction /i 2 . It makes an angle a 
with the floor, (a) How far up the ladder can a man climb before the ladder slips? (6) What is 
the condition that the ladder not slip at all regardless of where the man is located? 

Ans. (a) /j. 2 l(m + t&n a)/ (finx 2 + 1), (b) tana > l//t 2 

9.104. Work Problem 9.103 if the weight of the ladder is not 
negligible. 

9.105. A ladder AB of length I [Fig. 9-42] has one end A on an 
incline of angle a and the other end B on a vertical wall. 
The ladder is at rest and makes an angle /? with the incline. 
If the wall is smooth and the incline has coefficient of 
friction n, find the smallest value of n so that a man of 
weight W m will be able to climb the ladder without having it 
slip. Check your answer by obtaining the result of Problem 
9.37, page 241, as a special case. 




9.106. Work Problem 9.105 if the wall has coefficient of friction 



0i- 



Fig. 9-42 



A uniform rod AB with point A fixed rotates about a vertical axis so that it makes a constant 
angle a with the vertical [Fig. 9-43], If th e length of the rod is I, prove that the angular 
speed needed to do this is « = ^J(3g sec a)/2l . 



'//////////////////'. 






9.108. 



Fig. 9-43 



Fig. 9-44 



Fig. 9-45 



A circular cylinder of mass m and radius a is suspended from the ceiling by a wire as shown in 
Fig. 9-44. The cylinder is given an angular twist 6 and is then released. If the torque is assumed 
proportional to the angle through which the cylinder is turned and the constant of proportionality 
is X, prove that the cylinder will undergo simple harmonic motion with period 2iray/m/2\ . 

9.109. Find the period in Problem 9.108 if the cylinder is replaced by a sphere of radius a. 
Ans. 2jra\/2m/5X 

9.110. Work (a) Problem 9.108 and (6) Problem 9.109 if damping proportional to the instantaneous 
angular velocity is present. Discuss physically. 

9.111. A uniform beam AB of length I and weight W [Fig. 9-45] is supported by ropes AC and BD of 
lengths a and 6 respectively making angles a and /? with the ceiling CD to which the ropes 
are fixed. If equilibrium conditions prevail, find the tensions in the ropes. 



250 



PLANE MOTION OF RIGID BODIES 



[CHAP. 9 



9.112. In Fig. 9-46 the mass m is attached to a rope which is wound around 
a fixed pulley of mass M and radius of gyration K which can rotate 
freely about O. If the mass is released from rest, find (a) the angular 
speed of the pulley after time t and (b) the tension in the rope. 

9.113. Prove that the acceleration of the mass m in Problem 9.112 is 
ga 2 /(a 2 + K 2 ). 

9.114. Describe how Problem 9.112 can be used to determine the radius of 
gyration of a pulley. 



w////////////////. 



9.115. 



9.116. 



9.117. 



9.120. 



9.123. 




Fig. 9-46 



A uniform rod AB [Fig. 9-47] of length I and weight W having its 
ends on a frictionless wall OA and floor OB respectively, slides 
starting from rest when its foot B is at a distance d from O. Prove 
that the other end A will l eave the wall when the foot B is at a 
distance from O given by ^yoT^ + 4d 2 . 

A cylinder of mass 10 lb rotates about a fixed horizontal axis through 
its center and perpendicular to it. A rope wound around it carries 
a mass of 20 lb. Assuming that the mass starts from rest, find its 
speed after 5 seconds. Ans. 128 ft/sec 




Fig. 9-47 



What must be the length of a rod suspended from one end so that it will be a seconds pendulum 
on making small vibrations in a plane? Ans. 149 cm 



9.118. A solid sphere and a hollow sphere of the same radius both start from rest at the top of an 
inclined plane of angle a and roll without slipping down the incline. Which one gets to the 
bottom first? Explain. Ans. The solid sphere 

9.119. A compound pendulum of mass M and radius of gyration K about a horizontal axis is displaced 
so that it makes an angle O with the vertical and is then released. Prove that if the center of 
mass is at distance a from the axis, then the reaction force on the axis is given by 



Mg 



#2 + a 2 



y/([K 2 + 2a 2 ] cos* - a 2 cos ) 2 + (K 2 sin 0) 2 



A rectangular parallelepiped of sides a, b, and c is suspended vertically from the side of length a. 
Find the period of small oscillations. 



9.121. Find the least coefficient of friction needed to prevent the sliding of a circular hoop down an 
incline of angle a. Ans. § tan a 

9.122. Find the period of small vibrations of a rod of length I suspended vertically about a point %l 
from one end. 



9.124. 



9.125. 



A pulley system consists of two solid disks of radius r x and r 2 
respectively rigidly attached to each other and capable of rotating 
freely about a fixed horizontal axis through the center O. A weight 
W is suspended from a string wound around the smaller disk as shown 
in Fig. 9-48. If the radius of gyration of the pulley system is K 
and its weight is w, find (a) the angular acceleration with which the 
weight descends and (b) the tension in the string. 
Ans. (a) WgrJiWri + wK 2 ), (b) WwK 2 /(Wr\ + wK 2 ) 

A solid sphere of radius 6 rolls on the inside of a smooth hollow 
sphere of radius a. Prove that the period for small oscillations is 
given by 2vy/7(a — b)/5g . 

A thin circular solid plate of radius a is suspended vertically from a 
horizontal axis passing through a chord AB [see Fig. 9-49]. If it 
makes small oscillations about this axis, prove that the frequency 
of such oscillations is greatest when AB is at distance a/2 from 
the center. 





Fig. 9-49 



CHAP. 9] 



PLANE MOTION OF KIGID BODIES 



251 



9.126. A uniform rod of length 51 is suspended vertically from a string of length 21 which has its other 

end fixed. Prove that the normal frequencies for small oscillations in a plane are — \\— and 
1 [^ 2 ^ * 5l 



1_ 

2ir 



-y , and describe the normal modes. 



9.127. A uniform rod of mass m and length I is suspended from one of its ends. What is the minimum 
speed with which the other end should be hit so that it will describe a complete vertical circle? 



9.128. 



9.129. 



(a) If the bob of a simple pendulum is a uniform solid sphere of radius a rather than a point 
mass, prove that the period for small oscillations is 2iry/l/g + 2a 2 /5gl . 

(b) For what value of I is the period in (a) a minimum? 

A sphere of radius a and mass M rolls along a horizontal plane with constant speed v . It comes 
to an incline of angle a. Assuming that it rolls without slipping, how far up the incline will it 
travel? Ans. 10v§/(7flr sin a) 



9.130. Prove that the doughnut shaped solid or torus of Fig. 9-50 has 
a moment of inertia about its axis given by |Af(3a 2 + 4b 2 ). 

9.131. A cylinder of mass m and radius a rolls without slipping down 
a 45° inclined plane of mass M which is on a horizontal friction- 
less table. Prove that while the rolling takes place the incline 
will move with an acceleration given by mg/(SM + 2m). 

9.132. Work Problem 9.131 if the incline is of angle a. 
Ans. (mg sin 2a)/(3M + 2m - ra cos 2a) 



9.133. Find the (a) tension in the rope and (6) accelera- 
tion of the system shown in Fig. 9-51 if the radius 
of gyration of the pulley is 0.5 m and its mass 
is 20 kg. 




Fig. 9-50 



200 kg 



n = .2 



1\ 



9.134. Compare the result of Problem 9.133 with that 
obtained assuming the pulley to have negligible mass. 

9.135. Prove that if the net external torque about an axis 
is zero, then it is also zero about any other axis. 



a 

100 kg 



Fig. 9-51 



9.136. A solid cylindrical disk of radius a has a circular hole of radius b whose center is at distance c 
from the center of the disk. If the disk rolls down an inclined plane of angle a, find its acceleration. 
[See Fig. 9-52/ 

y 





Fig. 9-52 



Fig. 9-53 



9.137. Find the moment of inertia of the region bounded by the lemniscate r 2 = a 2 cos 2e [see Fig 9-531 
about the x axis. Ans. Ma 2 (3ir - 8)/48 

9.138. Find the largest angle of an inclined plane down which a solid cylinder will roll without slipping 
if the coefficient of friction is /t. 



9.139. Work Problem 9.138 for a solid sphere. 



252 



PLANE MOTION OF RIGID BODIES 



[CHAP. 9 



9.140. Discuss the motion of a hollow cylinder of inner radius a and outer radius b as it rolls down an 
inclined plane of angle a. 



9.141. A table top of negligible weight has the form of an equilateral triangle ABC of side s. The legs 
of the table are perpendicular to the table top at the vertices. A heavy weight W is placed on 
the table top at a point which is distant a from side BC and 6 from side AC. Find that part 
of the weight supported by the legs at A, B and C respectively. 



Ans. 



2Wa 2Wb 



Wll- 



2a + 26 



sV3 



9.142. Discuss the motion of the disk of Problem 9.136 down the inclined plane if the coefficient of 
friction is /<. 



9.143. A hill has a cross section in the form of a cycloid 

x = a{e + sin $), y — a(l — cos 9) 

as indicated in Fig. 9-54. A solid sphere of 
radius b starting from rest at the top of the hill 
is given a slight displacement so that it rolls 
without slipping down the hill. Find the speed 
of its center when it reaches the bottom of the hill. 




Ans. y/l0g(2a - b)/l 



Fig. 9-54 



9.144. Work Problem 3.108, page 85, if the masses and moments of inertia of the pulleys are taken 

into account. 

9.145. Work Problem 9.38, page 242, if friction is taken into account. 

9.146. A uniform rod of length I is placed upright on a table and then allowed to fall. Assuming that its 
point of contact with the table does not move, prove that its angula r velocity at the instant when 
it makes an angle d with the vertical is given in magnitude by y/3g{i--cose)/2l . 

9.147. Prove Theorem 9.11, page 228, for the case where the vibrations are not necessarily small. 
Compare Problem 9.41, page 243. 

9.148. A rigid body moves parallel to a given fixed plane. Prove that there is one and only one point 
of the rigid body where the instantaneous acceleration is zero. 

9.149. A solid hemisphere of radius a rests with its convex surface on a horizontal table. If it is 
displaced slightly, prove that it will undergo oscillations with period equal to that of a simple 
pendulum of equivalent length 4a/3. 



9.150. A solid cylinder of radius a and height h is suspended from axis AB 

as indicated in Fig. 9-55. Find the period of small oscillations about A 
this axis. 



9.151. Prove that a solid sphere will roll without slipping down an inclined 
plane of angle a if the coefficient of friction is at least f tan a. 

9.152. Find the least coefficient of friction for an inclined plane of angle a 
in order that a solid cylinder will roll down it without slipping. 

Ans. ^ tan a 



~«e*%%$mk 



B 




Fig. 9-55 



Chapter 10 



SPACE MOTION 
of RIGID BODIES 



GENERAL MOTION OF RIGID BODIES IN SPACE 

In Chapter 9 we specialized the motion of rigid bodies to one of translation of the 
center of -mass plus rotation about an axis through the center of mass and perpendicular 
to a fixed plane. In this chapter we treat the general motion of a rigid body in space. 
Such general motion is composed of a translation of a fixed point of the body [usually the 
center of mass] plus rotation about an axis through the fixed point which is not necessarily 
restricted in direction. 

DEGREES OF FREEDOM 

The number of degrees of freedom [see page 165] for the general motion of a rigid body 
in space is 6, i.e. 6 coordinates are needed to specify the motion. We usually choose 3 of 
these to be the coordinates of a point in the body [usually the center of mass] and the 
remaining 3 to be angles [for example, the Euler angles, page 257] which describe the 
rotation of the rigid body about the point. 

If a rigid body is constrained in any way, as for example by keeping one point fixed, 
the number of degrees of freedom is of course reduced accordingly. 

PURE ROTATION OF RIGID BODIES 

Since the general motion of a rigid body can also be expressed in terms of translation 
of a fixed point of the rigid body plus rotation of the rigid body about an axis through the 
point, it is natural for us to consider first the case of pure rotation and later to add the 
effects of translation. To do this we shall first assume that one point of the rigid body 
is fixed in space. The effects of translation are relatively easy to handle and can be obtained 
by using the result (10), page 167. 



VELOCITY AND ANGULAR VELOCITY OF 
A RIGID BODY WITH ONE POINT FIXED 

Suppose that point O of the rigid body % of Fig. 10-1 
is fixed. Then at a given instant of time the body 
will be rotating with angular velocity <■> about the in- 
stantaneous axis through O. A particle P of the body 
having position vector r„ with respect to O will have 
an instantaneous velocity v v given by 



— r v — »Xr» 



CO 



See Problem 10.2. 




Fig. 10-1 



253 



254 SPACE MOTION OF RIGID BODIES [CHAP. 10 

ANGULAR MOMENTUM 

The angular momentum of a rigid body with one point fixed about the instantaneous 
axis through the fixed point is given by 

= 2 ™< v {t v Xx v ) = 2 ™<v{rv X («» X r v )} {2) 

where m v is the mass of the vth particle and where the summation is taken over all 
particles of %. 

MOMENTS OF INERTIA. PRODUCTS OF INERTIA 

Let us choose a fixed xyz coordinate system having origin O and let us write 
a = a£ + a y j + nji, « = <*j. + o> y j + ajs. 
r v = Xvi + y v j + z v k 
Then equation (2) can be written in component form as [see Problem 10.3]. 



Q x = 7 *z w s + 7 xy<% + I xz"z 
9 -y = ty^x + hy% + hz^z 
Q z ~ hx^x + hy% + hz^z 



(•4) 



(6) 



where l xx = ^m^ + z*), I yy = Jm^ + 4 I zz = ^m^ + yl) (5) 

Ixy — ^ Wl v X v yv — ': 

lyz - - 2 m " V* z » - I 

lzx — — • ■ W/v Z11 Xv —- lxz 

The quantities hx, Iyy, Lz are called the moments of inertia about the x, y and z axes respec- 
tively. The quantities I xy , hz, ... are called products of inertia. For continuous mass 
distributions these can be computed by using integration. 

Note that the products of inertia in (6) have been defined with an associated minus 
sign. As a consequence minus signs are avoided in (4). 

MOMENT OF INERTIA MATRIX OR TENSOR 

The nine quantities I xx , I xy , ...,hz can be written in an array often called a matrix or 
tensor given by 

lxx lxy la 

lyx Iyy lyz I V / 

\lzx Izy lz 

and each quantity is called an element of the matrix or tensor. The diagonal consisting 
of the elements hxjyyjzz is called the principal or main diagonal. Since 

Ixy — lyx, Ixz — lzx, lyz = Izy \P) 

it is seen that the elements have symmetry about the main diagonal. For this reason (7) 
is often referred to as a symmetric matrix or tensor. 

KINETIC ENERGY OF ROTATION 

The kinetic energy of rotation is given by 

T = Wxx-l + hy< + hz-l + 2I xy<»x<»y + ^ xz^z + 2I yz<»y<»z) 

= i<o-0 {9) 



CHAP. 10] 



SPACE MOTION OF RIGID BODIES 



255 



PRINCIPAL AXES OF INERTIA 

A set of 3 mutually perpendicular axes having origin O which are fixed in the body 
and rotating with it and which are such that the products of inertia about them are zero, 
are called principal axes of inertia or briefly principal axes of the body. 

An important property of a principal axis [which can also be taken as a definition] 
is that if a rigid body rotates about it the direction of the angular momentum is the same 
as that of the angular velocity. Thus 

O = Jo, (10) 

where / is a scalar. From this we find [see Problem 10.6] that 

Vx + ^-^K + U = ° !► («) 



l zx<»x + l zy (A y + Vzz - 7 )° 



= 



In order that (11) have solutions other than the trivial one w x = 0, a y = 0, » z = 0, we 
require that 

ixx 1 Ixy *xz 

lyx Iyy-I hz = <> (12) 

Izx i-zy Izz L 

This leads to a cubic equation in / having 3 real roots h,h,h. These are called the 
principal moments of inertia. The directions of the principal axes can be found from (11), 
as shown in Problem 10.6 by finding the ratio o» x : w : «> 2 . 

An axis of symmetry of a rigid body will always be a principal axis. 



ANGULAR MOMENTUM AND KINETIC ENERGY ABOUT 
THE PRINCIPAL AXES 

If we call <a v <a 2 , (o 3 and Q lt Q 2 , o 3 the magnitudes of the angular velocities and angular 
momenta about the principal axes respectively, then 



°i — A* !* n 2 _7 2 a, 2' n 3 - h^s 



The kinetic energy of rotation about the principal axes is given by 

t = WA + *A + W) 

which can be written in vector form as [compare equation (9)] 



T = |a>-0 



(13) 

(U) 
(15) 



THE ELLIPSOID OF INERTIA 

Let n be a unit vector in the direction of <■>. Then 

o> = a>n = w(C0S a i + COS P j + COS y k) 



(16) 



where cos a, cos /3, cos y are the direction cosines of « or n with respect to the x, y and z 
axes. Then the kinetic energy of rotation is given by 



T = \h 



(17) 



256 



SPACE MOTION OF RIGID BODIES 



[CHAP. 10 



where I = I xx cos 2 a + I yy cos 2 ft + I zz cos 2 y 

+ 27 xy COS a COS /? + 27 ' COS /3 COS y + 27 xz COS a COS y 

By defining a vector 



p = n/yT 



where p = Px i + P J + P k, (18) becomes 



7 p 2 + 7 p 2 + 7 p 2 + 27 P p+ 27 P p+ 27 op =1 

xxrx yy"y zz"z xyrxry yzryrz zx"z"x 



(18) 
(19) 

(20) 



In the coordinates p x ,p y ,p z equation (20) represents an ellipsoid which is called the 
ellipsoid of inertia or the momental ellipsoid. 

If the coordinate axes are rotated to coincide with the principal axes of the ellipsoid, 
the equation becomes 

*A + h?\ + 1 A = 1 ( 21 ) 

where p v p 2 ,p 3 represent the coordinates of the new axes. 



EULER'S EQUATIONS OF MOTION 

It is convenient to describe the motion of a rigid body relative to a set of coordinate 
axes coinciding with the principal axes which are fixed in the body and thus rotate as the 
body rotates. If A lf A 2 , a 3 and c^, <o 2 , w 3 represent the respective components of the external 
torque and angular velocity along the principal axes, the equations of motion are given by 



Vl + (*s _ -^K^ = A l 

h^2 + (*1 ~ -^sK^l ~ A 2 
Vs + (^2 _ ^l) *^ ~ A 3 

These are often called Euler's equations. 



(22) 



FORCE FREE MOTION. 

THE INVARIABLE LINE AND PLANE 

Suppose that a rigid body is rotating about a 
fixed point O and that there are no forces acting 
on the body [except of course the reaction at the 
fixed point]. Then the total external torque is 
zero. Thus the angular momentum vector Q is 
constant and so has a fixed direction in space as 
indicated in Fig. 10-2. The line indicating this 
direction is called the invariable line. 

Since the kinetic energy is constant [see 
Problem 10.34], we have from (15) 



wo = constant 



(23) 




Fig. 10-2 



This means that the projection of » on O is constant, so that the terminal point of <* 
describes a plane. This plane is called the invariable plane. 

As the rigid body rotates, an observer fixed relative to the body coordinate axes would 
see a rotation or precession of the angular velocity vector a> about the angular momentum 
vector CI. 



CHAP. 10] 



SPACE MOTION OF RIGID BODIES 



257 



POINSOTS CONSTRUCTION. 
SPACE AND BODY CONES 

As noted by Poinsot, the above ideas can 
be geometrically interpreted as a rolling 
without slipping of the ellipsoid of inertia 
corresponding to the rigid body on the 
invariable plane. The curve described on the 
invariable plane by the point of contact with 
the ellipsoid is called the herpolhode [see 
Fig. 10-3]. The corresponding curve on the 
ellipsoid is called the polhode. 

To an observer fixed in space it would 
appear that the vector a traces out a cone 
which is called the space cone. To an 
observer fixed on the rigid body it would 
appear that <■> also traces out a cone which is 
called the body cone. The motion can then 
be equivalently described as a rolling without 
slipping of one cone on the other. See 
Problem 10.19. 



POLHODE. HERPOLHODE. 




Fig. 10-3 



SYMMETRIC RIGID BODIES. ROTATION OF THE EARTH 

Simplifications occur in the case of a symmetric rigid body. In such case at least 
two principal moments of inertia, say h and h, are equal and the ellipsoid of inertia 
is an ellipsoid of revolution. We can then show [see Problem 10.17] that the angular 
velocity vector <» precesses about the angular momentum vector O with frequency given by 



/ 



1 


h-h 


2tt 


h 



in) 



where the constant A is the component of the angular velocity in the direction of the 
axis of symmetry. 

In the case of the earth, which can be assumed to be an ellipsoid of revolution flattened 
slightly at the poles, this leads to a predicted precession period of about 300 days. In 
practice, however, the period is found to be about 430 days. The difference is explained as 
due to the fact that the earth is not perfectly rigid. 



THE EULER ANGLES 

In order to describe the rotation of a rigid 
body about a point we use 3 angular coordinates 
called Euler angles. These coordinates denoted 
by 4>, 9, xp are indicated in Fig. 10-4. In this 
figure the xyz coordinate system can be rotated 
into the x'y'z' system by successive rotations 
through the angles <£ and then 6 and then ^ 
[see Problem 10.20]. The line OA is sometimes 
called the line of nodes. 

In practice the x',y',z' axes are chosen as 
the principal axes or body axes of the rigid 
body while the x, y and z axes or space axes 
are fixed in space. 




Fig. 10-4 



258 



SPACE MOTION OF RIGID BODIES 



[CHAP. 10 



ANGULAR VELOCITY AND KINETIC ENERGY 
IN TERMS OF EULER ANGLES 



w l» w 2' w 3 



In terms of the Euler angles the components 
x f , y' and z' axes are given by 

<£ sin sin ^ + 6 sin ip 

$ sin 6 cos xp — 6 sin ^ 

<£ cos 8 + } 

The kinetic energy of rotation is then given by 

T = *(/,«? + V5 + W) 
where h, h, h are the principal moments of inertia. 



of the angular velocity along the 



(25) 



(26) 



MOTION OF A SPINNING TOP 

An interesting example of rigid body motion 
occurs when a symmetrical rigid body having one 
point on the symmetry axis fixed in space is set 
spinning in a gravitational field. One such example 
is that of a child's top as shown in Fig. 10-5, where 
point O is assumed as the fixed point. 

For a discussion of the various kinds of motion 
which can occur, see Problems 10.25-10.32 and 10.36. 




Fig. 10-5 



GYROSCOPES 

Suppose a circular disk having its axis mounted in gimbals [see Fig. 10-6] is given a 
spin of angular velocity a. If the outer gimbal is turned through an angle, the spin axis 
of the disk will tend to point in the same direction as previously [see Fig. 10-7]. This 
assumes of course that friction at the gimbal bearings is negligible. 

In general the direction of the spin axis remains fixed even when the outer gimbal, 
which is attached to some object, moves freely in space. Because of this property the 
mechanism, which is called a gyroscope, finds many applications in cases where maintaining 
direction [or following some specified course] is important, as for example in navigation 
and guidance or control of ships, airplanes, submarines, missiles, satellites or other moving 
vehicles. 

A gyroscope is another example of a symmetric spinning rigid body with one point on 
the symmetry axis [usually the center of mass] taken as fixed. 

Spin 





Fig. 10-6 



Fig. 10-7 



CHAP. 10] SPACE MOTION OF RIGID BODIES 259 

Solved Problems 

GENERAL MOTION OF RIGID BODIES IN SPACE 

10.1. Find the number of degrees of freedom for a rigid body which (a) can move freely 
in space, (b) has one point fixed, (c) has two points fixed. 

(a) 6 [see Problem 7.2(a), page 172] 

(6) 3 [see Problem 7.2(6), page 172] 

(c) If two points are fixed, then the rigid body rotates about the axis joining the two fixed 
points. Then the number of degrees of freedom is 1, such as for example the angle of rotation 
of the rigid body about this axis. 

10.2. A rigid body undergoes a rotation of angular velocity a about a fixed point O. Prove 
that the velocity v of any particle of the body having position vector r relative to O 
is given by v = «xr. 

This follows at once from Problem 6.1, page 147, on noting that the velocity relative to the 
moving system is dr/dt\ M = dr/dt\ b = 0. 

ANGULAR MOMENTUM. KINETIC ENERGY. 
MOMENTS AND PRODUCTS OF INERTIA 

10.3. Derive the equations (4), page 254, for the components of angular momentum in terms 
of the moments and products of inertia given by equations (5) and (6), page 254. 

The total angular momentum is given by 

N N 

G = 2 m„(r„ XvJ = 2 m v {r v X ( w X r„)} 

v=l v=l 

where we have used Problem 10.2 applied to the ?th particle. 
Now by equation (7), page 5, we have 

r„ X (« X r„) = «(r„ • r„) — r„(«* • r„) 

= (w x i + u„j + w x k)(*2 + y% + z v ) 

- {x v i + y v \ + z v k)(a x x v + Uy y v + u> z z v ) 

- WxiVv + z l) - UyXvVv - <*z x V Z V }i 

+ {a y («$ + z\) - UxXvVv ~ u z Vv z v}J 

+ {o> z (xl + y v ) — u x x v z v — u y y v z v }k 

Then multiplying by m„, summing over v and equating the coefficients of i, j and k to Sl x , S2 y and Sl z 
respectively, we find as required 

«x = ]2 ™ v (yl +z*)><* x + < — 2 m v x v y v > «„ + < — 2 m v x v z v 

— Ixx u x "f" •'xj/ w j/ + *xz u z 

Vy = ] 2 m v X v y v > OJ x + j 2 ™>v(xl + *v) f «y + \ ~ 2 ™>vVvZv \ ">x 

= Ixy u x + lyy^y ~^~ lyz^z 

- 2 nh*v*v f u x + "j ~ 2 rn^y^, > a y + -l 2 ™>v(tf + Vv)\<*z 

= *xz u x + lyz^y + *zz u z 

For continuous mass distributions of density a, we can obtain the same results by starting with 
O = | a(rX\)dr = | <x{r X (<• X r)} dr 



260 



SPACE MOTION OF RIGID BODIES 



[CHAP. 10 



10.4. If a rigid body with one point fixed rotates with angular velocity w and has angular 
momentum CI, prove that the kinetic energy is given by T = |<*> • CI. 

= |2m,{(»Xr,)-(«Xr v )} = £ 2 w„{« • [r„ X (. X r„)]} 
= £• • 2 wi„r„ X((.Xr„) = i tt -0 

N 

where we have used the abbreviation 2 in place of 2 • 



10.5. Prove that the kinetic energy in Problem 10.4 can be written 

T = W X A + I y y y + I Z A + 21 xM + 2I xz ^ z + 2I yz ^ z ) 



From Problem 10.4, we have 



T = l«> • CI 



i{«a;(4a; w x + 4i/ w y + 4z"«) 

+ Uyilyx^x + ^«/y w y + ^2/Z w z) 

+ w z(4x w a: + -fz2,«3, + 4z«z)} 

i(^xx w x + lyyUy + I Z z<*z + ^xy u x u y + ^xz a x u z + 2I yz w y a z ) 



using 



the fact that /«. = I vx , I xz = I zx , I. 



/„.. 



PRINCIPAL MOMENTS OF INERTIA AND PRINCIPAL AXES 

10.6. Derive equations (11), page 255, for the principal moments of inertia and the 
directions of the principal axes. 



Using CI = I* 

together with equations (3) and (-4), page 254, we have 



(1) 



= Ia y 
xz u x ~r *uz u u + 'z2« z — "* 



"■xy^y 



*yx u x + ^j/y w r/ + ^j/« w x: 



l yz>»y 



(hx ~ I)<*x + ^"j/ + 4« w z = 
lyx^x + Uj/j/ ~ I) u y "+" ^j/z w z = 
4z«x + lyz^y + (I zz ~ I)<*z — ° 



(2) 



The principal moments of inertia are found by setting the determinant of the coefficients of 
u x , o) y , u z in (2) equal to zero, i.e., 



* T.V. * 



*xy 
*yy ~ * 



'■xz 

*yz 

I,, -I 



= 



This is a cubic equation in / leading to three values I lt / 2> I3 which are the principal moments 
of inertia. By putting I = 7 X in (2) we obtain ratios for a x : u y : u z which yields the direction of » 
or the direction of the principal axis corresponding to I v Similarly, by substituting I 2 and I 3 
we find the directions of the corresponding principal axes. 



CHAP. 10] 



SPACE MOTION OF RIGID BODIES 



261 



10.7. Find the (a) moments of inertia and (b) products of inertia of a uniform square 
plate of length a about the x, y and z axes chosen as shown in Fig. 10-8. 

(a) The moment of inertia of an element dx dy of the plate 
about the x axis if the density is a is ay 2 dx dy. Then the 
moment of inertia of the entire plate about the x axis is jjj 



/a pa 
I ay 2 dx dy 



|a»4 = L M a 2 (1) 



since the mass of the plate is M — aa 2 . 

Similarly, the moment of inertia of the plate about 
the y axis is 

J id pa 
I a x 2 dxdy = i<ra 4 = ±Ma 2 (2) 

as is also evident by symmetry. 



f— u— - — dx dy 



Fig. 10-8 



The moment of inertia of dx dy about the z axis is a{x 2 + y 2 ) dx dy, and so the moment of 
inertia of the entire plate about the z axis is 

hz = f" f <r(x 2 + y 2 ) dx dy = %Ma 2 + %Ma 2 = \Ma 2 (3) 

This also follows from the perpendicular axes theorem [see page 226]. 

(6) The product of inertia of the element dx dy of the plate about the x and y axes is axydxdy, 
and so the product of inertia of the entire plate about these axes is 



= /,. 



pa pa 
— — I I axy dx dy — — j<t« 4 



-\Ma 2 

4 



(4) 



x=0 *'a = 



The product of inertia of the element dx dy of the plate about the * and z axes is the 
product of adxdy by the distances to the yz and xy planes, which are x and respectively. 
Thus we must have 



= /, 



0, 



and similarly 



hu = 



(5) 



%Ma 2 - I 


-\Ma 2 







-\Ma 2 


\Ma 2 - I 













|Ma 2 - 


- J 



10.8. Find the (a) principal moments of inertia and (b) the directions of the principal 
axes for the plate of Problem 10.7. 

(a) By Problem 10.6 and the results (l)-(5) of Problem 10.7, we obtain 



(1) 



or [(%Ma 2 - I)(±Ma 2 - I) - (-lMa 2 )(-lMa 2 )][%Ma 2 - I] = 

which can be written 

[P - |Ma2J + £-M 2 a*][%Ma 2 - 1} = 

Setting the first factor equal to zero and using the quadratic formula to solve for /, we 
find for the three roots of (1), 

I^-^Ma 2 , / 2 = JLMa2, 7 3 = \Ma 2 (2) 

which are the principal moments of inertia. 



(6) To find the direction of the principal axis corresponding to I lt we let I = I x = j^Ma 2 in 
the equations 

(±Ma 2 -I)a x - lMa 2 u y + <o 2 = 

-±Ma 2 a> x + (±Ma 2 -I)a y + <* z = [ (3) 

0<a x + Wy + (|Ma2-7)a> 2 = 



262 



SPACE MOTION OF RIGID BODIES 



[CHAP. 10 



The first two equations yield u y — w x while the third gives w x = 0. Thus the direction 
of the principal axis is the same as the direction of the angular velocity vector 

<■» = w x i + w y j + u z k — u x i + uj = u x (i + j) 

Then the principal axis corresponding to /j is in the direction i + j. 

Similarly, by letting I — I 2 ~ -^Ma 2 in (3) we find 
u y — —u x , u z = so that the direction of the correspond- 
ing principal axis is » = u x i — u x j = a x (i — j) or i — j. 

If we let I = I 3 = %Ma 2 in (3) we find a x = 0, 
u y = while u z is arbitrary. This gives <■> = to z k which 
shows that the third principal axis is in the direction k. 

The directions of the principal axes are indicated by 
i + j, i — j and k in Fig. 10-9. Note that these are 
mutually perpendicular and that i + j and i — j have the 
directions of the diagonals of the square plate which are 
lines of symmetry. 

The principal moments of inertia can also be deter- 
mined by recognizing the lines of symmetry. Fig. 10-9 




10.9. Find the principal moments of inertia at the center of a uniform rectangular plate 
of sides a and b. 

The principal axes lie along the directions of symmetry and thus must be along the x axis, 
y axis and z axis [the last of which is perpendicular to the xy plane] as in Fig. 10-10. 

By Problems 9.6, 9.9 and 9.11 the principal moments of inertia are found to be I t = ^Ma 2 , 
I 2 = ^-Mb 2 , I 3 = ^M(a 2 + b 2 ). 

y 



O 



dr = dz dy dx 



V* 2 + v 2 




Fig. 10-10 



Fig. 10-11 



10.10. Find the principal moments of inertia at the center of the ellipsoid 

,.2 „,2 ^2 

= 1 



— _|_ ^— 4- — 



a" 

One eighth of the ellipsoid is indicated in Fig. 10-11. The moment of inertia of the volume 
element dr of mass a dr about the z axis or "3" axis is (x 2 + y 2 )a dr, and the total moment of 
inertia about the z axis is 



yJl- x 2 /a 2 s*c\ll - (r 



r r x 



2/„2 , .,2,^2, 



(x 2 + y 2 )a dz dy dx 



x =o "y=0 



Integration with respect to z gives 

pa s>b\/l - x 2 /a 2 

8ac J J (* 2 + V 2 ) Vl - (aW + y 2 /b 2 ) dy dx 



x=0 y=0 



To perform this integration let x — aX, y — bY where X and Y are new variables. Then the 
integral can be written 

j (a 2 X 2 + b 2 Y 2 ) Vl - {X 2 + Y 2 ) dY dX 



CHAP. 10] 



SPACE MOTION OF RIGID BODIES 



263 



Introducing polar coordinates R, 6 in this XY plane, this becomes 

• 1 s*tt/2 

'R = *^0 = O 



} (a 2 R 2 cos 2 + 6 2 « 2 sin 2 6) Vl - R 2 R dR dQ 
= 2iraabc{cfi + b 2 ) | R^VT^WdR = ±7roabc(a 2 + b 2 ) 



R = 



where we use the substitution 1 — R 2 = t/ 2 in evaluating the last integral. 

Since the volume of the ellipsoid is ^-n-abc, the mass is M = ^Traabc and hence 7 3 
By symmetry we find /j = iM(6 2 + c 2 ), J 2 = iM(o 2 + c 2 ). 



LM(a 2 + ft 2 ). 



10.11. Suppose that the ellipsoid of Problem 10.10 is an oblate spheroid such that a = b 
while c differs slightly from a or b. Prove that to a high degree of approximation, 
(7 3 - 7i)//i = 1 - c/a. 



From Problem 10.10, if a — b then 



a 2 + c 2 



(a — c)(a + c) 
a 2 -|- c 2 



But if c differs 



only slightly from a then a + c « 2a and a 2 + c 2 « 2a 2 . Thus, approximately, 

(/ 3 - I x )/I = (a - c)(2a)/2a 2 = 1 - c/a 

10.12. Work Problem 10.11 for the case of the earth assumed to be an oblate spheroid. 

Since the polar diameter or distance between north and south poles is very nearly 7900 miles 
while the equatorial diameter is very nearly 7926 miles, then taking the polar axis as the "3" axis 
we have 2c = 7900, 2a = 7926 or c = 3950, a = 3963. 

Thus by Problem 10.11, (Z 3 - h)/I x = 1 ~ 3950/3963 = .00328. 



ELLIPSOID OF INERTIA 

10.13. Suppose that the moments and products of inertia of a rigid body % with respect 
to an xyz coordinate system intersecting at origin O are I xx , hv, Izz> I*y> Ixz, hz respec- 
tively. Prove that the moment of inertia of % about an axis making angles a, p, y 
with the x, y and z axes respectively is given by 

7 = Ixx COS 2 a + I yy COS 2 P + Izz COS 2 y 

+ 21 xy COS a COS P + 21 xz COS a COS y + 21 yz COS p COS y 



by 



A unit vector in the direction of the axis is given 

n = cos a i + cos )3 j + cos Y k 

Then if m„ has position vector r„, its moment of inertia 
about the axis OA is m v T)1 where D v - |r„ X n|. But 




and 



i J k 

r„Xn = x v y v z v 

cos a cos p cos y 

= (y v cos y — z v cos /3)i 

+ (z v cos a — x v cos y)j 

+ (x v cos /3 — y v cos a)k Fig. 10-12 

\ Tv x n| 2 = (y v cos y — z v cos (3) 2 + (z v cos a - x v cos y) 2 + (*„ cos fi - y v cos a) 2 

= (vl + 4) cos 2 a + (x 2 + 4) cos 2 £ + (x 2 v + yl) cos 2 y 

- 2x v y v cos a cos (3 - 2x„z„ cos a cos y — 2y v z v cos /? cos y 



264 SPACE MOTION OF RIGID BODIES [CHAP. 10 

Thus the total moment of inertia of all masses ra„ is 

= J2 ™v(vl + 4) I co S 2 a + | 2 m^ajj! + 4) \ cos 2 /? + 1 2 m^ 2 . + yl) I cos 2 y 
+ 2 -j - 2 myX v V v Y cos a cos + 2 -j — 2 VA f cos « cos y 



+ 2^-2 m v y v z v > cos /? cos y 

= I xx cos 2 a + lyy cos 2 /? + /^ cos 2 y 

+ 2/j.j, cos a cos /? + 2/ X2 cos a cos y + 2I yz cos /? cos y 

10.14. Find an equation for the ellipsoid of inertia corresponding to the square plate of 
Problem 10.7. 

We have from Problem 10.7, 

I xx = \Ma\ l m = ±Ma 2 , I zz = *Ma*, I xy = -i-Ma 2 , I xz = 0, /„, = 
Then the equation of the ellipsoid of inertia is by equation (20), page 256, 
\MaVx + iMaVy + \Mtfpl - \Ma? 9xPy = 1 

or ?\ + P% + 2 P l - h x Py = 3/Ma* 

EULER'S EQUATIONS OF MOTION 

10.15. Find a relationship between the time rate of change of angular momentum of a rigid 
body relative to axes fixed in space and in the body respectively. 

If the rigid body axes are chosen as principal axes having directions of the unit vectors 
e lt e 2 and e 3 respectively, then the angular momentum becomes 

CI = ^U)^ + / 2 «2 e 2 + ^3"3C3 

Now by Problem 6.1, page 147, if s and b refer to space (fixed) and body (moving) axes 
respectively, then 

at Is dt \b 

= hu^! + 7 2 « 2 e 2 + h^3 e 3 

+ ( W 1©1 + "2^2 + "3^3) X (/l"iei + I 2 ^2^2 + h w aPs) 

— { Jl"l + (^3 ~ ■^2) w 2' d 3} e l + {-^2"2 + (^ 1 ~~ ^3)"l«3}e2 
+ {h^3 + (1 2 ~ ^l) w 2 w l}e 3 

10.16. Derive Euler's equations of motion (22), page 256. 

By the principle of angular momentum, we have 

A = Tt\. W 

where A is the total external torque. Writing 

A = A^ + A 2 e 2 + A 3 e 3 (2) 

where A 1? A 2 , A 3 are the components of the external torque along the principal axes and making 
use of (1) and Problem 10.15, we find 



/jwj + (I 3 — / 2 )u 2 w 3 = A t 
7 2 w 2 + (h ~ h)^3 u i — A 2 
7 3 w 3 + (/ 2 — /i)wi<o 2 = A 3 



(3) 



CHAP. 101 



SPACE MOTION OF RIGID BODIES 



265 



FORCE FREE MOTION OF A RIGID BODY. ROTATION OF THE EARTH 

10.17. A rigid body which is symmetric about an axis has one point fixed on this axis. 
Discuss the rotational morion of the body, assuming that there are no forces acting 
other than the reaction force at the fixed point. 

Choose the axis of symmetry coincident with one of the principal axes, say the one having 
direction e 3 . Then I t = 7 2 and Euler's equations become 



A"i + (^3 "~ 7i)o> 2 o> 3 — 

/i<o 2 + (Ii — I 3 )w 3 ui = 

/ 3 «3 = 
From (3), o> 3 = constant - A so that (1) and (2) become after dividing by I lt 

/ 7 S - /j \ 

«i + f — j )Aw 2 = 

h ~ h 



0) 2 + 



Aw, = 



Differentiating (5) with respect to t and using (U), we find 

•73-/^2 



+ 



h 

o> 2 ~^~ k2 <°2 



A 2 w 2 = 



where 



73-^ 



h 



Solving (7), we find o> 2 = B cos xt + C sin Kt 

If we choose the time scale so that w 2 = when t = 0, then 

w 2 = C sin Kt 

Then from (5) we have u t = C cos Kt 

Thus the angular velocity is 

a = wjej + w 2 e 2 + 10363 

= C cos Kt e x + C sin Kt e 2 + Ae s (11) 

From this it follows that the angular velocity is 
constant in magnitude equal to w = |«| = VC 2 + A 2 
and precesses around the "3" axis with frequency 

\h -7/ 



(1) 

(2) 

U) 
(5) 

(6) 

(7) 
(8) 



(9) 
(10) 



f 2,7 



2tt7! 



^A 



(12) 



as indicated in Fig. 10-13. 

Note that the vector <■> describes a cone about the 
"3" axis. However, this motion is relative to the 
body principal axes which are in turn rotating in 
space with angular velocity u. 




Fig. 10-13 



10.18. Calculate the precession frequency of Problem 10.17 in the case of the earth rotating 
about its axis. 

Since the earth rotates about its axis once in a day, we have <o 3 = A = 2w radians/day. Then 
the precessional frequency is from Problems 10.12 and 10.17, 

/ = ~( 3 7 1 A = 77-(l--M = IT- (.00328)(2*-) = .00328 radians/day 
cir \ l 1 J &tt \ Ci/ Zir 

The period of precession is thus P = 1// = 305 days. The actual observed period is about 430 days 
and is explained as due to the non-rigidity of the earth. 



266 



SPACE MOTION OF RIGID BODIES 



[CHAP. 10 



THE INVARIABLE LINE AND PLANE. 

POLHODE, HERPOLHODE, SPACE AND BODY CONES 

10.19. Describe the rotation of the earth about its axis in terms of the space and body cones. 
From Problem 10.17 the angular velocity o> and angular momentum CI are given respectively by 

o) = w 1 e 1 + w 2 e 2 ~^~ w 3 e 3 = C(cos Kt ej + sin Kt e 2 ) + Ae 3 

CI = 7 1 « 1 e 1 + 7 1 w 2 e2 + ^3 w 3 e 3 = hC ( cos Kt e i + sm Kt e 2) + 7 3 Ae 3 
Let a be the angle between o> 3 = w 3 e 3 = Ae 3 and CI. Then 



and 



« 3 -0 = |« 3 ||0|cosa = A V/f C 2 + ZfA 2 cos a = 7 3 A2 

7 3 A 



V/fC 2 + 7 2 A 2 

Similarly, let /? be the angle between » 3 and m. Then 

» 3 • a> = |«* 3 | |<a| cos/3 = Ay/C 2 + A 2 cos /3 

and cos /? 



(1) 



A 2 



From (1) and (2) we see that 



7 X C 



Thus 



V/fC 2 + 7fA 2 
tan « = 

tan a 



V^ + A2 



sin/8 = 



VC 2 + A2 



I S A 



tan /3 



Q 

tan ft = — 
A 

h 
h 



<*) 

(5) 



Now for the earth or any oblate spheroid [flattened at the poles] we have 7 X < 7 3 . It follows 
that a < (3. 

The situation can be described geometrically by Fig. 10-14. The cone with axis in the direction 
CI is fixed in space and is called the space cone. The cone with axis <* 3 = u 3 e 3 is considered as 
fixed in the earth and is called the body cone. The body cone rolls on the space cone so that the 
element in common is the angular velocity vector <■>. Now 



<l) 3 X <!> 



ei e 2 e 3 

A 

I^C cos Kt I t C sin Kt I 3 A 



= — A7 X C sin Kt e 1 + A7 X C cos Kt e 2 



Thus 



CI • (o> 3 X ») 



(I^C cos Kt ej + 7jC sin Kt e 2 + 7 3 Ae 3 ) 

• (— A7 t C sin Kt ex + AI X C cos/ct e 2 ) 




It follows from Problem 1.21(6), page 16, that CI, o> 3 and o> 
lie in one plane. 

An observer fixed in space would see the vector <o 
tracing out the space cone [see Fig. 10-14]. An observer 
fixed in the body would see the vector « tracing out the body 
cone. 

In the case of the earth the space cone is inside the 
body cone due to the fact that l x < 7 3 . For the case of a 
prolate spheroid the reverse situation is true, i.e. 7 t > 7 3 and 
the space cone is outside the body cone [see Problem 10.121]. 




Cone 



Fig. 10-14 



CHAP. 10] 



SPACE MOTION OF RIGID BODIES 



267 



THE EULER ANGLES 

10.20. Show by using three separate figures how the xyz coordinate system of Fig. 10-4, 
page 257, is rotated into the x'y'z' coordinate system by successive rotations through 
the Euler angles <j>, 6 and ^. 



Refer to Figures 10-15, 10-16 and 10-17. Fig. 
10-15 indicates a rotation through angle <f> of the x and y 
axes into an X and Y axis respectively while keeping 
the same z or Z axis. 

In Fig. 10-16 a rotation about the X axis through 
angle e is indicated so that the Y and Z axes of 
Fig. 10-15 are carried into the Y' and Z' axes of 
Fig. 10-16 respectively. 

In Fig. 10-17 a rotation about the Z' or z' axis 
through angle ^ is indicated so that the X' and Y' axes 
are carried into the x' and y' axes respectively. 

In the figures we have indicated unit vectors on the 
x, y, z axes; X, Y, Z axes; X' , Y' , Z' axes and x', y', z' 
axes by i,j,k; I, J, K; I', J', K' and i', j',k' respectively. 



z or Z 




Fig. 10-15 




Z' or z' 



XorX' 




Fig. 10-16 



Fig. 10-17 



10.21. Find the relationships between the unit vectors (a) i,j, k and I, J, K of Fig. 10-15, 
(b) I, J, K and F, J', K' of Fig. 10-16, (c) I', J', K' and i', j', k' of Fig. 10-17. 



(a) From Fig. 10-15, 

i = (i-I)I + (i-J)J + (i-K)K 
j = (j • 1)1 + (j-J)J + (j-K)K 
k = (k-I)I + (k-J)J + (k-K)K 

(6) From Fig. 10-16, 

I = (I-I')I' + (I-J')J' + (I'K')K' 
J = (J'I')r + (J-J')J' + (J-K')K' 
K = (K'I')I' + (K'J')J' + (K«K')K' 

(c) From Fig. 10-17, 

F = (I'-i')i' + (I'-j')J' + (I'-k')k' 

J' = (J'-i')i' + (J'«j')j' + (J'-k')k' 

K' = (K'«i')i' + (K'»j')j' + (K'-k')k' 



cos I — sin <f> J 
sin <f> I + cos J 
K 



= I' 



cos e J' — sin e K' 
sin e J' + cos e K' 



cos \p i — sin \p j 
sin ^ i' + cos ^ j' 
k' 



268 SPACE MOTION OF RIGID BODIES [CHAP. 10 

10.22. Express the unit vectors i, j,k in terms of i', j', k\ 

From Problem 10.21, 

i = cos I — sin <p 3, j = sin I + cos <p 3, k = K 

I — V, J = cos e 3' — sin K', K - sin 3' + cos K' 

I' = cos0i' — sin0j', J' = sin i' + cos0j', K' — k' 

Then i — cos I — sin 3 = cos <f> V — sin <p cos J' + sin <p sin K' 

= cos cos i' — cos sin j' 

— sin cos sin i' — sin cos cos j' + sin sin k' 

= (cos cos — sin cos sin 0)i' 

+ (— cos sin — sin cos cos 0)j' + sin sin k' 

j = s in I + cos J = sin I' + cos cos 3' — cos sin K' 

— sin cos i' — sin sin j' 

+ cos cos (9 sin i' + cos cos cos 3' — cos sin * k' 

— (sin cos + cos cos sin 0)i' 

+ (— sin sin + cos cos cos 0)3' — cos sin k' 

k — sin J' + cos JK' = sin sin i' + sin cos 3' + cos k' 



10.23. Derive equations (25), page 258. 

o> = co^k + Wfl I' + c^K' = 0k + Si' + 0K' 

= sin (9 sin i' + sin cos 3' + cos k' 
+ cos i' — sin 3' + 0k' 

= (0 sin (9 sin + cos 0)i' 

+ (0 sin cos — sin 0)3' + (0 cos + 0)k' 

Then since o> = u x \' + w a 'j' + co 2 'k', 

« x ' = w t = sin sin + cos 

<V — "2 — sm * cos ~~ sin 

<o s ' = <o 3 = cos + 

10.24. (a) Write the kinetic energy of rotation of a rigid body with respect to the principal 
axes in terms of the Euler angles, (b) What does the result in (a) become if h = hi 

(a) Using Problem 10.23, the required kinetic energy is seen to be 

t = £(/ lW ? + 1 2 4 + 1 A) 

= ^/j(0 sin sin + cos 0) 2 

+ -1-/2(0 sin cos - sin 0) 2 + i/ 3 (0 cos + 0) 2 

(6) If /j = 7 2 , the result can be written 

T = i/ x (02 sin 2 + 02) + 1/3(0 cos + 0) 2 

MOTION OF SPINNING TOPS AND GYROSCOPES 

10.25. Set up equations for the motion of a spinning top having fixed point O [see Fig. 10-18]. 



CHAP. 10] 



SPACE MOTION OF RIGID BODIES 



269 



Let xyz represent an inertial or fixed set 
of axes having origin O. Let x'y'z' represent 
principal axes of the top having the same 
origin. Choose the orientation of the x'y' 
plane so that Oz, Oz' and Oy' are coplanar. 
Then the x' axis is in the xy plane. The 
line ON in the x'y' plane making an angle ^ 
with the x' axis is assumed to be attached 
to the top. 

The angular velocity corresponding to 
the rotation of the x'y'z' axes with respect to 
the xyz axes is 



w — wie^ + w 2 e 2 ~^~ w 3 e 3 



(1) 



In obtaining the angular momentum we must 
use the fact that in addition to the com- 
ponent w 3 due to rotation of the x'y'z' system 
there is also the component s = se 3 = ^e 3 
since the top is spinning about the z' axis. 
Then the angular momentum is 




Fig. 10-18 



O — /xwje! + 7 2 <o 2 e 2 + h^z + s ) e 3 



(*) 



Now if we let subscripts / and b denote the fixed system and body system respectively, we have 
by Problem 6.1, page 147, 

dCl 



dt 



<ZO I 

/ dt \b 



(3) 



Using (1) and (2) in (3), we find 

~~fo L = ih^i + (h ~ h)^2^3 + h u 2 s )^i 

+ ih^2 + (^1 — ^3) w l u 3 ~ ^3 w l s ) e 2 

+ {-^("3 + *) + (h — -^l)"l"2} e 3 

The total torque about O is 

A = (fc g ) X (mg) = (le s ) x (-mgk) 



Since 



the torque is 
Then using A = 



k = (k • e 1 )e 1 + (k • e 2 )e 2 + (k • e 3 )e 3 = cos (ir/2 — fl)e 2 + cos 6 e 3 
= sin e 2 + cos e 3 

A = —mgl(e 3 X k) 
dCl 



dt If 



mgl sin e e x 
with I t — I 2 , we find from (4) and (6), 



-^l w l + (^3 — A) W 2 W 3 + ^3"2 S = m 9l sm * 
^1«2 + (^1 — ^3)wi"3 — ^3<0lS = 

7 3 (<u 3 + s) = 



U) 



(5) 



(6) 



(7) 



10.26. Express equations (7) of Problem 10.25 in terms of the Euler angles 6 and <£ of 
Fig. 10-18. 

The components o> v « 2 , w 3 can be obtained from Problem 10.23 by formally letting ^ = 0. 



We find 

Ul — 0, u 2 = sm 9, a5 3 = COS 6 

Then equations (7) of Problem 10.25 become 



(1) 



270 SPACE MOTION OF RIGID BODIES [CHAP. 10 

h '°' + (^3 — h)$ 2 sm e cos e + h$ s sm 6 = m ^ sin e 
hC<P sin<9 + $e cose) + (7j — I 3 )e^>cose — I 3 es =0 V (2) 

7 3 ( cos o — $e sin e + s) — 

The quantities <£, and s are often known as the magnitudes of the angular velocity of precession, 
of nutation and of spin respectively. 

10.27. Prove that the equations (2) of Problem 10.26 can be written as 

(a) he - 7i<J) 2 sin cos + Z" 3 A</>sin0 = mgl sin e 

(b) h{'4 sin e + 2j>6 cose) - hA8 = 
where A is a constant. 

From the third equation in (7) of Problem 10.25, 

w 3 + s = A or s = A — « 3 (1) 

Then substitution into the first and second of equations (7) yields 

7i"i ~ 7 1 w 2 w3 + 7 3 « 2 A — m 9l sin e W 

/ 1 6j 2 + 7 1 <o 1 « 3 — /3W1A = W 

Using the results (I) of Problem 10.26, we find that equations (2) and (3) reduce to the required 
equations. 

10.28. (a) Find the condition for steady precession of a top. 

(b) Show that two precessional frequencies are possible. 

Since is constant so that V = 0, we have from Problem 10.27(a), 

(Irf 2 cos e — 7 3 A<£ + mgl) sin = 
or Irf 2 cos — I 3 A$ + mgl = 

7 3 A ± V7fA 2 - Amglh cos e 
from which <t> = g^-^ I > 

Thus there are two frequencies provided that 

/fA 2 > Amglli cos e (2) 

If 7 3 A 2 = 4mgll x cos only one frequency is possible. 

If A is very large, e.g. if the spin of the top is very great, then there are two frequencies, 
one large and one small, given by 

I 3 A/(I lC ose), mgl/I 3 A W 

10.29. Prove that 

(a) %h(e 2 + <J» 2 sin 2 8) + |/ 3 A 2 + mgl cos e - constant = E 

(b) h4> sin 2 d + hA cos e = constant = K 
and give a physical interpretation of each result. 

(a) Multiply equations (7) of Problem 10.25 by Wl , co 2 and <* 3 + s respectively, and add to obtain 

/ifttji! + *>2«2) + 7 3("3 + *)(«S +«) = m ^ Sin ' * 

which can be written as 

f t {#M + 4) + W a * + s > 2 > = f t (~ m 9lcose) 
Integrating and using <* 3 + s = A as well as the results Wl = * and <o 2 = sine, we find 
ll^e* + <£ 2 sin 2 0) + |7 3 A 2 + mgl cos e = E CO 



CHAP. 10] SPACE MOTION OF RIGID BODIES 271 

where E is constant. The result is equivalent to the principle of conservation of energy, 
since the kinetic energy is 

T = AJ^fl 2 + ^ 2 sin2 e) + %I 3 A* (2) 

while the potential energy is V = mgl cos e (8) 

and T + V — E is the total energy. 

(6) Multiplying the result of Problem 10.27(6) by sin e, 

I^sirfo + 11 x <t>6 sin e cos e — I 3 Ae sine = 

which can be written 

jt(/i0 sin 2 + I 3 A costf) = 

Integrating, Irf sin 2 + I 3 A cos 6 — constant = K (-4) 

To interpret this result physically, we note that the vertical component of the angular 
momentum is Irf sin 2 e + I 3 A cos e [see Problem 10.123], and this must be constant since the 
torque due to the weight of the top has zero component in the vertical direction. 



10.30. Let u = cos 6. Prove that: 

(a) u 2 = {a-/3u)(l-u 2 )- {y-Su) 2 = f(u) 

where a = 2(E - ihA 2 )/h, p = 2mgl/h, y = K/I u 8 = hA/h; 



(6) t = j 



du 

7m 



+ constant 



(a) From Problem 10.29, 



i/ 1 (ff 2 + <p 2 sin 2 e) + ±I 3 A* + mgl cos 6 = E (1) 

Irf sin 2 6 + I 3 A cos = K {2) 



K — I 3 A cos e 
From (2), * = / lSin 2, <*> 

Substituting this into (I), 

(K - I 3 A cos 0)2 
i' 1 ' 2 + 2 / lS in2 g + ¥* A + mgU ° se = E 

Letting u = cos * so that u = —sin 8 h and sin 2 9 = 1 — u 2 , this becomes 

y.r^ + 27.(1-^) + "*"" = £ -* 7 ^ 2 
Thus * + (^p) 2 + 2rog '^-" 2) = «&^a v _ 1 / 1 A, 

which can be written as 

w 2 = (a - fiu)(l - u*) - (y - 8u) 2 = /(«) (4) 

where a = (2tf - I^A 2 )/^, p = 2mgl/I 1 , y = K/I lt 8 = /aA/^ (5) 

Note that with this notation {3) can be written 

. _ y - 8u . . 

* ~ T^ 2 " (<?) 



272 



SPACE MOTION OF RIGID BODIES 



[CHAP. 10 



(b) From the result of (a) we have, since u > 0, 






Integrating, 



£ = 



s 



or 



+ c 



<&* 



(7) 



The integral can be evaluated in terms of elliptic functions which are periodic. 



10.31. (a) Prove that 6 = at those values of m for which 

f(u) = («-/8M)(l-M 2 )-(y-8M) 2 - 

(b) Prove that the equation in (a) has three real roots u u u 2 ,u 3 but that in general 
not all the angles corresponding to these are real. 



(a) From Problem 10.30(a), u 2 = f(u) = (a - £m)(1 ~ « 2 ) - (y - 8u) 2 

Since u = -sin e 0, it follows that e = where m = or /(«) = 0. 

roots of the equation 

/( M ) = ( a - fiu)(l - m 2 ) - (y - S«)2 = 

(b) Equation (1) can be written as 

f(u) = /3tt 3 - (S 2 + a)u 2 + (2yS - /3)u + a - y 2 



(1) 

Thus 6 = at the 

(2) 



Since /? > 0, it follows that 

/(I) 



= -(y-S) 2 , /(-l) 



Thus there is a change of sign from 
— to + as u goes from 1 to °°, and 
consequently there must be a root, say 
u 3 , between 1 and » as indicated in 
Fig. 10-19. 

Now we know that in order for 
the motion of the top to take place we 
must have f(u) = u 2 s o. Also, since 
^ e ^ ir/2, we must have = tt = 1. 
It thus follows that there must be two 
roots m x and u 2 between and 1, as 
indicated in the figure. 

It follows that in general there are 
two corresponding angles e x and e 2 such 



that cos 0i 



cos 2 — u 2- ^ n special 



cases it could happen that u x 

U<l = Ug = 1. 




Fig. 10-19 



10.32. Give a physical interpretation of the results found in Problem 10.31. 

The fact that there are two roots % and u 2 corresponding to 6 X and 2 respectively, shows 
that the motion of the top is such that its axis always makes an angle e with the vertical which 
lies between e y and e 2 . This motion, which is a bobbing up and down of the axis between the 
limits <?! and 2 , is called nutation and takes place at the same time as the precessional motion 
of the axis of the top about the vertical and the spinning of the top about its axis. Because 
the motion can be expressed in terms of elliptic functions [see Problem 10.104], we can show 
that it is periodic. 

In general the tip of the axis of the top will describe one of various types of curves such as 
indicated in Figs. 10-20, 10-21 and 10-22. The type of curve will depend on the root of the 
equation [see equation (6) of Problem 10.30] 



<t> = 



_ y — Su 
- u 2 



(1) 



CHAP. 10] 



SPACE MOTION OF RIGID BODIES 



273 



If this root given by y/8 is greater than u 2 , the curve of Fig. 10-20 occurs. If it is the same as 
u 2 , the curve of Fig. 10-21 obtains. If it is between u x and u 2 the curve of Fig. 10-22 occurs. Other 
cases can arise if the root is the same as u x or is less than u x [see Problems 10.124]. 




y/8 > u 2 
Fig. 10-20 




y/8 = u 2 
Fig. 10-21 




u-l < y/8 < u 2 
Fig. 10-22 



Aside from the general motion which is made up of nutation and precession, there are 
various special cases which can arise. One of these is the case of steady precession with no 
nutation [see Problem 10.28]. In this case u x = u 2 so that e x ~ e 2 or e = constant. Another 
case is the "sleeping top" which occurs where u 2 = u 3 = 1 and the axis of the spinning top always 
remains vertical [see Problem 10.36]. 



MISCELLANEOUS PROBLEMS 

10.33. If T is the total kinetic energy of rotation of a rigid body with one point fixed, 
prove that dT/dt - <o • A where all quantities refer to the body principal axes. 

Multiplying both sides of the Euler equations (3) of Problem 10.16 by w 1 ,w 2 , <o 3 respectively 
and adding, we obtain 

^l w l w l ~^~ •^2 (0 2 w 2 ~f~ ^3 W 3 W 3 = "l-^l ~^~ c0 2-^2 ~^~ w 3-^3 (7) 



But 
and 

Thus (1) becomes 



t • ^ t • lt • _l^/r2j_r2.r 2 \— dT 
iiCOjWj + i 2 O) 2 0) 2 + '3 W 3 W 3 — o di lWl 2< ° 2 3< ° 3 ' — ~dt 

0)^ + 0) 2 A 2 + "3-^3 = ("l^l + «2 e 2 + <">3 e 3) * ( A l e l + A 2 e 2 + A 3 e 3 ) 
= 6> • A 



dT/dt 



at • A 



(2) 
(3) 



10.34. (a) Prove that if there are no forces acting on a rigid body with one point fixed, 
then the total kinetic energy of rotation is constant, (b) Thus prove that © • O = 
IT — constant. 

(a) Since there are no forces, A = 0. Then by Problem 10.33, dT/dt - or T = constant. 
(6) Since O = /jo^ej + I 2 u 2 e 2 + 730)363 and «• = o^e^ + o> 2 e 2 + 0)363, 

«* • O = / x o)f + 7 2 w| + / 3 o>3 = 2T — constant 

10.35. Find the precession frequency of Problem 10.17 in terms of the kinetic energy and 
angular momentum of the rigid body. 

The kinetic energy is 

T = i(/,o) 2 + 7 2 o)| + I 8 «|) = 4(/ lW ? + h4 + 7 3W |) = ¥h° 2 + J s A2 ) 



so that 



/jC 2 + 7 3 A2 = 2T 



(1) 



274 



SPACE MOTION OF RIGID BODIES 



[CHAP. 10 



The angular momentum is 

O = Il03 1 e 1 + /2 w 2 e 2 + ^3 w 3 e 3 = ^l w l e l + ^l"2 e 2 + ^3 w 3 e 3 

= 7 1 C(cos Kt e l + sin id e 2 ) + ^3-Ae 3 

so that = |Q| = V/fC 2 + /|A2 or 

I 2 C 2 + 1%A 2 = S22 

Solving (i) and (2) simultaneously, we find 

277 3 - fi2 fi 2 _ 277, 

(72 - A 2 - 

hda-It)' A h(h-h) 

Then from Problem 10.17, equation (12), the precession frequency is 



2^/i \ 



\(n*-2Ti 1 )(i 3 -i l )\ 



(2) 



(3) 



04) 



10.36. Find the condition for a "sleeping top". 

For a "sleeping top" we must have e = and 5 = 0, since the axis must remain vertical 
and no nutation can take place. Then from Problem 10.29, 



hA 



K, 



I 3 A 2 - 2(E-mgl) 

Also, from Problem 10.30 we have a = 2mgl/I l , = 2mgl/I v y = l 3 All x , 8 = I 3 A/I V Thus a - ($ 
and y = 8, and so 

/(h) = (a - j8u)(l - h2) - (y - 8h)2 = «(1 - u)(l - it*) - y 2 (l - u)2 = (1 - h)2 [«(1 + u) - y2] 
It follows that /(h) = has a double root at u — 1, while the third root is given by 

v 2 



1 = 



J|A« 



- 1 



2mgll 1 

Then the top will "sleep" if this root is greater than or at most equal to 1, so that 

A2 ^ Amglljll 

Of course, even though this condition may apply at the beginning, energy will in practice 
be diminished due to friction at the support so that after some time we will have A 2 < Amglljl^. 
In such case precession combined with nutation will be introduced. Further loss of energy will 
ultimately cause the top to fall down. 



10.37. Find the torque needed to rotate a rectangular plate of sides a and b [see Fig. 10-23] 
about a diagonal with constant angular velocity ©. 

By Problem 10.9 the principal moments of inertia of the 
plate at the center O are given by 



l x = ^M<fi t I 2 = ^Mb*, I z = -^M(a 2 + b*) 

We have 

a> = (wi)i 4- (»*j)j 



Thus 

Substituting (1) and 



Va 2 + 6 2 



Va 2 + 6 2 



(1) 



(2) 



— osb 



w 2 



V» 2 + 6 2 ' 



Va 2 + 6 2 ' 
into Euler's equations 

/jWi + (7 3 — 72) W 2 W 3 = A l 
I 2 a 2 + (/ x — 73)<°3 W 1 = A 2 
/ 3 <J) 3 + (J 2 — Ji)"i"2 = A 3 




Fig. 10-23 



w 3 



CHAP. 10] 



SPACE MOTION OF RIGID BODIES 



275 



M(b 2 — a z )abu 2 „,, , . , A , ^ ^ . 

we find Aj = 0, A 2 = 0, A 3 = U( 2 + b*) ■ Thus the re( l uired torque about O is 

M(&2 - cfi)ab<J- 
A ~ 12(a2 + 62) 

Note that if the rectangular plate is a square, i.e. if a — b, then A = 0. 



(4) 



Supplementary Problems 



GENERAL MOTION OF RIGID BODIES IN SPACE 

10.38. Find the number of degrees of freedom of (a) a sphere free to roll on a plane, (b) an ellipsoid 
free to rotate about a fixed point, (c) an airplane moving in space. 

Ans. (a) 3, (b) 3, (c) 6 

10.39. In Fig. 10-24 a displacement of a tetrahedron in 
space is indicated. Show directly that the dis- 
placement can be accomplished by a translation 
plus a rotation about a suitable axis, thus 
illustrating Chasle's theorem [page 224] for 
space. 

10.40. Give an illustration similar to that of Problem 
10.39 involving a rigid body whose surfaces are 
not plane surfaces. 

10.41. Derive the result of Problem 10.2, page 259, 
without using Problem 6.1, page 147. 




Fig. 10-24 



ANGULAR MOMENTUM. KINETIC ENERGY. MOMENTS AND PRODUCTS OF INERTIA 

10.42. A rigid body consists of 3 particles of masses 2,1,4 located at (1,-1,1), (2,0,2), (—1,1,0) respec- 
tively. Find the angular momentum of the body if it is rotated about the origin with angular 
velocity » = 3i - 2j + 4k. Arts. -6j + 42k 

10.43. Determine the (a) moments of inertia about the x, y and z axes and (b) the products of inertia 
for the rigid body of Problem 10.42. 

Ans. (a) I xx = 12, I yy = 16, I„ = 16; (6) I xy = 6, I yz = 2, 1„ = -6 



10.44. What is the kinetic energy of rotation for the system of Problem 10.42? 



Ans. 180 



10.45. Find the (a) moments of inertia and (6) products of inertia of a uniform rectangular plate 
ABCD of sides AB = a and AD — b taken about axes AB, AD and the line perpendicular to 
the plate at B. 
Ans. (a) I xx = iM62, Iyy = l M a?, I gz = ±M(a? + 62) 

(&) hy = ~lMab, l yz = 0, I xz = 
calling axes through AB and AD the x and y axes respectively. 



10.46. Find the (a) moments of inertia and (6) products of inertia of a cube of side a taken about x,y, z 
axes coinciding with three intersecting edges of the cube. 

Ans. (a) I xx = I yw = I zz - %Ma2, (b) I xy = I yz = I xz - -\Ma* 

10.47. Find the (a) angular momentum and (6) kinetic energy of rotation of the cube of Problem 10.46 
about the point of intersection O of the three edges if the cube has an angular velocity 
a, = 2i + 5j - 3k about O. Ans. (a) ^MaHlOi + 43j - 45k), (6) 185MaV12 




276 SPACE MOTION OF RIGID BODIES [CHAP. 10 

10.48. Find the (a) moments of inertia and (b) products of inertia of the uniform solid sphere 
x 2 + y 2 + z 2 = a 2 in the first octant, i.e. in the region *g0, 2/ = 0, z ^ 0. 

Ans. (a) I xx = I m = I zz = §Ma 2 , (b) I xy = I yz = I xz = -2MaV5ir 

PRINCIPAL MOMENTS OF INERTIA. PRINCIPAL AXES. ELLIPSOID OF INERTIA 

10.49. Prove that the principal moments of inertia for a system consisting of two particles of masses 
m l and m 2 connected by a massless rigid rod of length I are I t = J 2 — m l m 2 ^/{m l + m 2 ), I z = 0. 

10.50. Find the (a) principal moments of inertia and (b) directions of the principal axes for the system 
of Problem 10.42. 

Ans. (a) I x = 18, I 2 = 13 - t/T3 , I 3 = 13 + \/73 

(b) j + k, £(/ + V73)i-J + k, j(i_-/73)i_j + k> 

10.51. Determine the (a) principal moments of inertia and 
(b) directions of the principal axes for right triangle ABC 
of Fig. 10-25 about point C. 

10.52. Find the principal moments of inertia at the center of a 
parallelogram of sides a and b and acute angle a. 

10.53. Find the (a) principal moments of inertia and (6) direc- 
tions of the principal axes for the cube of Problem 10.46. 
Ans. (a) Zj = 7 2 = flilfa 2 , I 3 = ±Ma 2 

(6) Axis associated with I s is in the direction of the diagonal from the origin. Axes 
associated with J x and I 2 have any mutually perpendicular directions in a plane perpen- 
dicular to this diagonal. 

10.54. Find the principal moments of inertia of a uniform cylinder of radius a and height h. 
Ans. J a = I 2 = jLM(3a 2 + h 2 ), I 3 = \Ma 2 

10.55. Obtain the principal moments of inertia and directions of principal axes for a rectangle of 
sides a and b by using Problem 10.45 and equations (11), page 255. Compare with Problem 10.9, 
page 262. 

10.56. Find the lengths of the axes of the ellipsoid of inertia corresponding to the rectangle of 
Problem 10.55. Ans. 4^3/Ma 2 , Ay/Z/Mb 2 , 4^S/M(a 2 + b 2 ) 

10.57. Find the lengths of the axes of the ellipsoid of inertia corresponding to the cube of Problem 10.46. 
Ans. 4\/3/llMa 2 , 4V3/llMa 2 , 2 v / 6/Ma 2 

10.58. Prove that the ellipsoid of inertia for a regular tetrahedron is a sphere and determine its radius. 

10.59. If I V I 2 ,I 3 are the principal moments of inertia, prove that 

I, ^ I 2 + h, h ^ h + h. h ^ h + h 

10.60. Under what conditions do any or all of the equality signs hold in Problem 10.59? 

10.61. Prove that if a rigid body is a solid of revolution about a line L, then L is a principal axis 
corresponding to any part of L. 

10.62. Suppose that a rigid body is symmetrical about a plane P. Prove that if L is a line perpendicular 
to P at point O, then L is a principal axis corresponding to point O. 

EULER'S EQUATIONS OF MOTION 

10.63. A rigid body having one point O fixed and no external torque about O, has two equal principal 
axes of inertia. Prove that it must rotate with constant angular velocity. 



CHAP. 10] SPACE MOTION OF RIGID BODIES 277 

10.64. Write Euler's equations for the case of plane motion of a rigid body and discuss their physical 
significance. 

10.65. Solve the problem of a compound pendulum by using Euler's equations. 

10.66. Describe how Euler's equations can be used to discuss the motion of a solid cylinder rolling down 
an inclined plane. 

10.67. Write Euler's equations in case the axes are not principal axes. 

FORCE FREE MOTION. INVARIABLE LINE AND PLANE. 
POLHODE, HERPOLHODE, SPACE AND BODY CONES 

10.68. If two principal moments of inertia corresponding to the fixed point about which a rigid body 
rotates are equal, prove that (a) Poinsot's ellipsoid is an ellipsoid of revolution, (6) the polhode 
is a circle and (c) the herpolhode is a circle. 

10.69. Discuss the (a) invariable line and plane, (6) polhode and herpolhode and (c) space and body 
cones for the case of a rigid body which moves parallel to a given plane, i.e. plane motion of a 
rigid body. 

10.70. (a) How would you define the instantaneous axis of rotation for space motion of a rigid body? 
(6) What is the relationship between the instantaneous axis of rotation and the space and body 
cones? 

10.71. Prove that relative to its center of mass the axis about which the earth spins in a day will 
rotate about an axis inclined at 23.5° with respect to it in 25,780 years. 

THE EULER ANGLES 

10.72. Using the notation of Problem 10.20, page 267, find: (a) I, J,K in terms of i,j,k; (6) I\J',K' in 
terms of I,J,K; (c) i',j',k' in terms of I',J',K'. 

Ans. (a) I = cos <f> i + sin <f> j, J = —sin $ i + cos <f> j, K = k 

(6) I' = I, J' = cos e J + sin e K, K' = —sin e J + cos e K 
(c) i' = cos ^ F + sin ^ J', J' = —sin f I' + cos $ J', k' = K' 

10.73. Prove the results u x = 9 cos <f> + ^ sin e sin <j> 

a y = sin <f> — ^ sin e cos 
u g = $ + yp cos *• 

10.74. If I x = I 2 = I3, prove that the kinetic energy of rotation of a rigid body referred to principal 
axes is T = \I X (4? + o 2 + j< 2 + 2w cos e). 

MOTION OF SPINNING TOPS AND GYROSCOPES 

10.75. A top having radius of gyration about its axis equal to 6 cm is spun about its axis. The spinning 
point is fixed and the center of gravity is on the axis at a distance 3 cm from this fixed point. 
If it is observed that the top precesses about the vertical at 20 revolutions per minute, find the 
angular speed of the top about its axis. Ans. 3.10 rev/sec or 19.5 rad/sec 

10.76. A uniform solid right circular cone of radius a and height h is spun so that its vertex is fixed 
and its axis is inclined at a constant angle a with the vertical. If the axis precesses about the 
vertical with period P, determine the angular speed of the cone about its axis. 

10.77. Work Problem 10.76 if the cone is surmounted by a uniform solid hemisphere of radius a and the 
same density. 

10.78. Explain physically why the spin axis of the gyroscope of Figures 10-6 and 10-7, page 258, 
should maintain its direction. 



278 



SPACE MOTION OF RIGID BODIES 



[CHAP. 10 



10.79. Explain how a gyroscope can be used to enable a ship, airplane, submarine or missile to follow 
some specified course of motion. 



MISCELLANEOUS PROBLEMS 

10.80. A uniform solid cube of side a and mass M has its edges lying on the positive x, y and z axes 
of a coordinate system with vertex at the origin O. If it rotates about the z axis with constant 
angular velocity w, find the angular momentum. Arts, — J^Ma 2 <o(3i + 3j — 8k) 

10.81. Find the moment of inertia of a uniform solid cone of radius a, height h and mass M about 
(a) the base, (b) the vertex. Arts, (a) ^Ma 2 , (b) -^M{2h 2 + a 2 ) 

10.82. Find the principal moments of inertia at the center of a uniform elliptical plate having semi-major 
axis a and semi-minor axis b. Ans. I t = %Mb 2 , I 2 = \Ma 2 , I 3 = \M{a 2 + b 2 ) 

10.83. A top has the form of a solid circular disk of radius o and mass M 
with a thin rod of mass m and length I attached to its center 
[see Fig. 10-26]. Find the angular velocity with which the top 
should be spun so as to "sleep". Assume that the base point O 
is fixed. 

10.84. Work Problem 10.83 for a cone of radius a, height h and mass M. 

10.85. Work Problem 10.83 for a cone of radius a, height h and mass M 
surmounted by a hemisphere of radius a and mass m. 

10.86. A coin of radius a is set spinning about a vertical axis with 
angular velocity o> [see Fig. 10-27]. Prove that the motion is stable 
if to 2 > Agla. 

10.87. Suppose that the coin of Problem 10.86 is spun with angular speed 
s about a diameter which is inclined at an angle a with the vertical 
and which is fixed at point O. Assuming there is no nutation, 
find the angular speed with which the coin precesses about the 
vertical. 

10.88. Discuss how gyroscopes can be used to control the motions of a ship 
on a stormy sea. 




Fig. 10-26 




Fig. 10-27 



10.89. The vertex of a uniform solid cone of radius a, height h and mass M is fixed at point O of a 

horizontal plane. Prove that if the cone rolls on the plane with angular velocity <o about an axis 

, . . ..... 3Mh 2 (a? + 6fc 2 )co 2 

perpendicular to the plane through O, then the kinetic energy of rotation is 40(a 2 + h 2 ) ' 

10.90. Explain how the principal axes of a rigid body can be found if the direction of one of the 
principal axes is known. 

10.91. A uniform solid cone has the radius of its base equal to twice 
its altitude. Prove that the ellipsoid of inertia corresponding 
to its vertex is a sphere. 

10.92. Explain how a gyroscope can be used as a compass, often 
called a gyrocompass. 

10.93. A dumbbell consists of two equal masses M attached to a 
rod ABC of length I and negligible mass [see Fig. 10-28]. 
The system rotates about a vertical axis DCE with constant 
angular velocity «* such that the rod makes a constant 
angle 8 with the vertical. Prove that the angular momentum 
Q of the system describes a cone of angle irl2 — e about u 
and has magnitude \Ml 2 u sin e . 




CHAP. 10] 



SPACE MOTION OF RIGID BODIES 



279 



10.94. (a) Prove that the magnitude of the torque needed to keep the system of Problem 10.93 in motion 
is \Ml 2 v> 2 sin2<? - (&) What is the direction of the torque? 

10.95. Work (a) Problem 10.93 and (6) Problem 10.94 if the rod ACB has mass m. 

10.96. A thin solid uniform circular plate of radius a has its center 
attached to the top of a thin fixed vertical rod OA [see 
Fig. 10-29]. It is spun with constant angular speed w about 
an axis which is inclined at angle a with the normal OB to 
the plate, (a) Prove that the angular velocity vector «* 
precesses about the normal OB with period 2tt/(u cos a). 
(b) Prove that the axis OB describes a space cone with period 
2tt/(<o \/1 + 3 cos 2 a) . 

10.97. In Problem 10.96 find the angle through which the plate 
turns during the time it takes OB to describe the space cone. 

10.98. Find the principal moments of inertia of a uniform solid cone 
of radius a, height h and mass M taken about the (a) vertex, 
(6) center of mass. 
Ans. (a) I x = h = ^M(a 2 + Ah 2 ), I 3 = ^Ma 2 

(b) h=h= i>M(h* + 4a 2 ), 7 3 = J^Ma 2 




Fig. 10-29 



10.99. A compound pendulum of mass M oscillates about a horizontal axis which makes angles a, /?, y 
with respect to the principal axes of inertia. If the principal moments of inertia are 7 1 ,7 2 , 7 3 
respectively and the distance from the center of mass to the axis of rotation is I, prove that for 
small oscillations the period is 2v^Mgl/I where I = Ml 2 + 1^ cos 2 a + I 2 cos 2 (3 + / 3 cos 2 y. 

10.100. Find the period of small oscillations of a uniform solid 
cone which rotates about a horizontal axis attached to 
the vertex of the cone. 

10.101. An elliptical plate [see Fig. 10-30] having semi-major 
and semi-minor axes of lengths a and 6 respectively is 
rotated with constant angular speed « about an axis 
making a constant angle a with the major axis. Find 
the torque required to produce this motion. 

10.102. Work Problem 10.101 if the elliptical plate is replaced 
by an ellipsoid. Fig. 10-30 




10.103. Given Euler's equations of motion for a rigid body having zero external torque about a fixed 
point O, i.e., 

I^X + (/ 3 — jT 2 )u 2 <o 3 = 0, 7 2 «2 + Cl ~" -^3) W 3 W 1 = 0, Z 3 « 3 + (/ 2 — ^l)«l"2 = 



prove that 
and 



^i w i ~^~ ^2 W 2 ~t" ^3 U 3 = constant — 2T 
/?«i + /|u| + 7 3 <o| = constant = H 2 

10.104. Prove from Problem 10.103 that a lt « 2 and « 3 satisfy a differential equation of the form 
dy/dx = V(l — # 2 )(1 ~ k 2 x 2 ), and thus show that the angular velocity can be expressed in terms 
of elliptic functions. 

10.105. Find the moment of inertia of a uniform solid cone of radius a, height h and mass M about a 
line which lies in its surface. Ans. ^Ma 2 (a 2 + 6h 2 )/(a 2 + h 2 ) 

10.106. The moments and products of inertia of a rigid body about the x, y and z axes are I xx = 3, 
I yy = 10/3, I zz = 8/3, I xy = 4/3, I ^ = — 4/3, I yz = 0. Find (a) the principal moments of inertia and 
(6) the directions of the principal axes. 

Ans. (a) I x = 3, J 2 = 2, 7 3 = 4 

(6) ei = i - 2j - 2k, e 2 = -2i + j - 2k, e 3 = -2i - 2j + k 



280 



SPACE MOTION OF RIGID BODIES 



[CHAP. 10 



10.107. A cone having semi-vertical angle o rolls with constant angular speed w on a horizontal plane with 
its vertex fixed at a point O. Prove that the axis of the cone rotates about the vertical axis 
through O with angular speed w tan a. 

10.108. A horizontal plane rotates about a vertical axis with constant angular velocity <■». A uniform 
solid sphere of radius a is placed on this plane. Prove that the center describes a circle with 
angular velocity given in magnitude by |-w. 

10.109. Work Problem 10.108 if the sphere is not necessarily of constant density. 
Ans. aK 2 /(K 2 + a 2 ) where K is the radius of gyration about a diameter 

10.110. Show how to find the relative maximum and minimum distances from the origin to the ellipsoid 
* = Ax 2 + By 2 + Cz 2 + Dxy + Eyz + Fxz - 1. 

[Hint. Maximize or minimize the function * = x 2 + y 2 + z z subject to the condition * = 1. 
To do this use the method of Lagrange multipliers, i.e. consider the function G = * + X* where 
X is the (constant) Lagrange multiplier and set dG/dx, dG/dy, dG/dz equal to zero.] 

10.111. Explain the relationship of Problem 10.110 to the method of page 255 for obtaining principal 
moments of inertia and directions of principal axes. 

10.112. (a) Find the relative maximum and minimum distances from the origin to the ellipsoid 

9x 2 + 10y 2 + Sz 2 + Axy - 4scz = 3. 
(b) Discuss the connection of the results of (a) with those of Problem 10.106. 

10.113. Find the moment of inertia of the system of particles of Problem 10.42 about a line through the 
point (2, — 1, 3) in the direction 3i — 2j + 4k. 



10.114. Prove that the motion of the "sleeping top" of Problem 10.36 is stable if A 2 ^ Amglljl^ 

10.115. Find the moment of inertia of the lemniscate r 2 - a 2 cos 2e about the z axis 
10.116 



Ans. jfMa 2 



A plane rigid body (lamina) has an xy and x'y' coordinate system with common origin O such 
that the angle between the x and x' axes is a [see Fig. 10-31]. Prove that 
(a) I x > x > = I xx cos 2 a — 2I xy sin a cos a + I yy sin 2 a 



I rT sin 2 a + 21 rv sin a cos a + /„„ sin 2 a 




10.117. Use Problem 10.116 to prove that 
and give a physical interpretation. 

10.118. Referring to Problem 10.116, find an expression for I x - y - in terms of J xx ,I xy ,I yy and a. 

10.119. Use the results of Problems 10.116 and 10.118 to prove that for a plane region having moments 
and products of inertia defined by I xx , I xy> I yy corresponding to a particular xy coordinate system, 
the principal axes are obtained by a rotation of these axes through an angle a given by 
tan 2a = I xy /(I yy — I xx ). 

10.120. Prove that the lengths of the principal axes in Problem 10.116 are given by 



CHAP. 10] SPACE MOTION OF RIGID BODIES 281 

10.121. Discuss Problem 10.19, page 266, if l x > / 3 . 

10.122. Find the moment of inertia of a uniform semicircular wire of mass M and radius a about its 
center. Ans. 2M(ir — 2)a 2 /v 

10.123. Prove that the expression on the left side of equation (4) in Problem 10.29 is the vertical 
component of the angular momentum. 

10.124. Discuss Problem 10.32 if the root of equation (1) is (a) equal to u lf (b) less than u v 

10.125. A rigid body consists of 3 particles of masses m u ra 2 and ra 3 . The distance between m l and m 2 ; 

m 2 and ra 3 ; m s and m l are l 12 , hz an d Z 3 i respectively. Prove that the moment of inertia of the 

system about an axis perpendicular to the plane of the particles through their center of mass is 

given by .2 , ,2 , ,2 

Wi»»2*i2 + m 2 m 3 l 2 3 + m^m^ 

mi + m 2 + m 3 

10.126. Derive a "parallel axis theorem" for products of inertia and illustrate by means of an example. 

10.127. Prove that the principal moments of inertia of a triangle of sides a,b,c and mass M about the 
center of mass are given by 

I t = I 2 = ^(«2 + fe2 + C 2 ± 2 Va 4 + 6 4 + C4 - tt 2&2 _ fc2 c 2 _ C 2 a 2 ), / 3 = ff (« 2 + & 2 + C 2 ) 

10.128. A coin of radius 1.5 cm rolls without slipping on a horizontal table such that the plane of the 
coin makes an angle of 60° with the table. If the center of the coin moves at a speed of 3 m/sec, 
prove that the coin moves in a circular path and find its radius. Ans. 2.5 m 



Chapter 11 LAGRANGE'S 

EQUATIONS 



GENERAL METHODS OF MECHANICS 

Up to now we have dealt primarily with the formulation of problems in mechanics by 
Newton's laws of motion. It is possible to give treatments of mechanics from rather 
general viewpoints, in particular those due to Lagrange and Hamilton. 

Although such treatments reduce to Newton's laws, they are characterized not only 
by the relative ease with which many problems can be formulated and solved but by their 
relationship in both theory and application to such advanced fields as quantum mechanics, 
statistical mechanics, celestial mechanics and electrodynamics. 

GENERALIZED COORDINATES 

Suppose that a particle or a system of N particles moves subject to possible constraints, 
as for example a particle moving along a circular wire or a rigid body moving along an 
inclined plane. Then there will be a minimum number of independent coordinates needed 
to specify the motion. These coordinates denoted by 

q\, <?2, . . .,q n (1) 

are called generalized coordinates and can be distances, angles or quantities relating to 
them. The number n of generalized coordinates is the number of degrees of freedom 
[see page 165]. 

Many sets of generalized coordinates may be possible in a given problem, but a 
strategic choice can simplify the analysis considerably. 

NOTATION 

In the following the subscript a will range from 1 to n, the number of degrees of 
freedom, while the subscript v will range from 1 to N, the number of particles in the 
system. 

TRANSFORMATION EQUATIONS 

Let r„ = x v i + Vvj + z v k be the position vector of the vth particle with respect to an 
xyz coordinate system. The relationships of the generalized coordinates (1) to the 1 position 
coordinates are given by the transformation equations 

X v = X v (qi, #2, • • • , Qn, t) 

ijv = Vv{qi, q%, . . . , qn, t) [ (2) 

z v = z v (qi, q 2 , . . ., qn, t) 

where t denotes the time. In vector form, (2) can be written 

r v - r v {qi, q%, . . . , q n , t) (#) 

The functions in (2) or (3) are supposed to be continuous and to have continuous derivatives. 

282 



CHAP. Ill 



LAGRANGE'S EQUATIONS 283 



CLASSIFICATION OF MECHANICAL SYSTEMS 

Mechanical systems can be classified according as they are scleronomic or rheonomic, 
holonomic or non-holonomic, and conservative or non-conservative as defined below. 

SCLERONOMIC AND RHEONOMIC SYSTEMS 

In many mechanical systems of importance the time t does not enter explicitly in the 
equations (2) or (3). Such systems are sometimes called scleronomic. In others, as for 
example those involving moving constraints, the time t does enter explicitly. Such systems 
are called rheonomic. 

HOLONOMIC AND NON-HOLONOMIC SYSTEMS 

Let q it q 2 , ...,q n denote the generalized coordinates describing a system and let t 
denote the time. If all the constraints of the system can be expressed as equations having 
the form <j>{qi, q 2 , . . . , q n , t) = or their equivalent, then the system is said to be holonomic; 
otherwise the system is said to be non-holonomic. Compare page 170. 

CONSERVATIVE AND NON-CONSERVATIVE SYSTEMS 

If all forces acting on a system of particles are derivable from a potential function 
[or potential energy] V, then the system is called conservative, otherwise it is non-con- 
servative. 

KINETIC ENERGY. GENERALIZED VELOCITIES 

The total kinetic energy of the system is 

T = I 5>„r2 (*) 

6 v=l 

The kinetic energy can be written as a quadratic form in the generalized velocities q a . 
If the system is scleronomic [i.e. independent of time * explicitly], then the quadratic form 
has only terms of the form a^haq&. If it is rheonomic, linear terms in q a are also present. 

GENERALIZED FORCES 

If W is the total work done on a system of particles by forces F„ acting on the vth 
particle, then B 

dW = *i£$adq a (5) 



<x = l 

N 



dr v 



where *« = 2 Fv * T^~ ^ 



v = l 



dq a 



is called the generalized force associated with the generalized coordinate q a . See 
Problem 11.6. 

LAGRANGE'S EQUATIONS 

The generalized force can be related to the kinetic energy by the equations [see 

Problem 11.10] 

±(9T) -iT = * 8 (7) 

dt\dq a J dq a 



284 LAGRANGE'S EQUATIONS [CHAP. 11 

If the system is conservative so that the forces are derivable from a potential or potential 
energy V, we can write (7) as 

d(BL\ _ BL = 

dt\dq a J dq a 

where L = T - V (9) 

is called the Lagrangian function of the system, or simply the Lagrangian. 

The equations (7) or (8) are called Lagrange's equations and are valid for holonomic 
systems which may be scleronomic or rheonomic. 

If some of the forces in a system are conservative so as to be derivable from a potential 
V while other forces such as friction, etc., are non-conservative, we can write Lagrange's 
equations as 

*(*£) - *k = *; (10) 

dt\dq a J dq a 

where L-T-V and $>L are the generalized forces associated with the non-conservative 
forces in the system. 



GENERALIZED MOMENTA 

We define at 

dq a 
to be the generalized momentum associated with the generalized coordinate q a . We often 
call p a the momentum conjugate to q a , or the conjugate momentum. 

If the system is conservative with potential energy depending only on the generalized 
coordinates, then (11) can be written in terms of the Lagrangian L = T - V as 

BL 

Pa 



Bq a 



(12) 



LAGRANGE'S EQUATIONS FOR NON-HOLONOMIC SYSTEMS 

Suppose that there are m equations of constraint having the form 

VAadqa + Adt = 0, ^Badq a + Bdt = 0, ... (13) 

a a 

or equivalents ^A a q a + A = 0, ^B a q a + B = 0, ... (U) 

a « 

We must of course have m < n where n is the number of coordinates q a . 

The equations (13) or (U) may or may not be integrable so as to obtain a relationship 
involving the q a 's. If they are not integrable the constraints are non-holonomic or non- 
integrable; otherwise they are holonomic or integrable. 

In either case Lagrange's equations can be replaced by 

d_ (dT\ _ aT = ^ + XiAa + XzBa + ... (i 5) 

dt\dq a J dq a 
where the m parameters Ai, A 2 , ... are called Lagrange multipliers [see Problem 11.18]. 
If the forces are conservative, (15) can be written in terms of the Lagrangian 

L = T-V as * , , _ x _ r 

A 9L\ _dL_ = XiAa + X2Ba + ... (16) 

dt\dq a J dq a 



CHAP. Ill 



LAGRANGE'S EQUATIONS 



285 



It should be emphasized that the above results are applicable to holonomic (as well as 
non-holonomic) systems since a constraint condition of the form 

<l>{Ql,Q2, ...,Qn,t) = (17) 

can by differentiation be written as 

^p-d Qa + &dt = (18) 

which has the form (13). 



(19) 



LAGRANGE'S EQUATIONS WITH IMPULSIVE FORCES 

Suppose that the forces F„ acting on a system are such that 

lim C F v dt = Sv 

where t represents a time interval. Then we call F„ impulsive forces and $ v are called 
impulses. 

If we let the subscripts 1 and 2 denote respectively quantities before and after appli- 
cation of the impulsive forces, Lagrange's equations become [see Problem 11.23] 

'H.) - (2L) = f a 
^dq a /2 \dq a /i 



where 



fa — 2* *Jv 



dq a 



(20) 
(21) 



If we call f a the generalized impulse, (20) can be written 

Generalized impulse = change in generalized momentum 
which is a generalization of Theorem 2.6, page 36. 



(22) 



Solved Problems 



GENERALIZED COORDINATES AND TRANSFORMATION EQUATIONS 

11.1. Give a set of generalized coordinates needed to completely specify the motion of 
each of the following: (a) a particle constrained to move on an ellipse, (b) a circular 
cylinder rolling down an inclined plane, (c) the two masses in a double pendulum 
[Fig. 11-3] constrained to move in a plane. 

(a) Let the ellipse be chosen in the xy plane of Fig. 11-1. The particle of mass m moving on the 
ellipse has coordinates (x,y). However, since we have the transformation equations x = a cos 6, 
y = 6 sin e, we can specify the motion completely by use of the generalized coordinate 0. 





Fig. 11-1 



Fig. 11-2 




286 LAGRANGE'S EQUATIONS [CHAP. 11 

(b) The position of the cylinder [Fig. 11-2 above] on the inclined plane can be completely specified 
by giving the distance x traveled by the center of mass and the angle e of rotation turned 
through by the cylinder about its axis. 

If there is no slipping, x is related to e so that only one generalized coordinate [either x or e] 
is needed. If there is slipping, two generalized coordinates x and e are needed. 

(c) Two coordinates &i and 6 2 completely specify the positions of masses m l and w 2 [see Fig. 11-3 
above] and can be considered as the required generalized coordinates. 

11.2. Write the transformation equations for the system in Problem 11.1(c). 

Choose an xy coordinate system as shown in Fig. 11-3. Let {x x ,y^) and {x^y^t be the rectangular 
coordinates of m 1 and ra 2 respectively. Then from Fig. 11-3 we see that 

x x = l-i cos $i 2/i = l\ sin $i 

x 2 = h cos e i + l 2 cos 2 V2 ~ h sm * l + h sin 2 

which are the required transformation equations. 

11.3. Prove that |t = ^. 

dq<x dq a 

We have r„ = r v (q u q 2 , ..., q n > t). Then 

Thus * = £■ (•) 

dq a dq a 

We can look upon this result as a "cancellation of the dots". 

11.4. Prove that ±(—) = !r- 

dt\dqj 3 #« 

We have from (1) of Problem 11.3, 

dr v d 2 r v . d 2 r„ . fl2r v 

oq a dq a d<li d <la d <ln dq a dt 

A (Oil) - JL { ?IL) -!± 4- d ( dr v\ d( ln d ( ?>*v 

N ° W dt\dqj ~ d qi \dq a J dt + '" dq n \d qoi J dt -dt\dq a 

d*r v . d 2r v . d2r„ 

' «i + " " * + a~ z~ «« + IT^T W 



dtfl^« dq n dq<* dtdq a 

Since r„ is assumed to have continuous second order partial derivatives, the order of differ- 
entiation does not matter. Thus from (2) and (3) the required result follows. 
The result can be interpreted as an interchange of order of the operators, i.e., 

A(jL\ = — {A\ 

CLASSIFICATION OF MECHANICAL SYSTEMS 

11.5. Classify each of the following according as they are (i) scleronomic or rheonomic, 
(ii) holonomic or non-holonomic and (iii) conservative or non-conservative. 

(a) A sphere rolling down from the top of a fixed sphere. 



CHAP. Ill 



LAGRANGE'S EQUATIONS 287 



(b) A cylinder rolling without slipping down a rough inclined plane of angle a. 

(c) A particle sliding down the inner surface, with coefficient of friction ^, of a 
paraboloid of revolution having its axis vertical and vertex downward. 

(d) A particle moving on a very long frictionless wire which rotates with constant 
angular speed about a horizontal axis. 

(a) scleronomic [equations do not involve time t explicitly] 
non-holonomic [since rolling sphere leaves the fixed sphere at some point] 
conservative [gravitational force acting is derivable from a potential] 

(b) scleronomic 

holonomic [equation of constraint is that of a line or plane] 

conservative 

(c) scleronomic 
holonomic 

non-conservative [since force due to friction is not derivable from a potential] 

(d) rheonomic [constraint involves time t explicitly] 

holonomic [equation of constraint is that of a line which involves t explicitly] 

conservative 



WORK, KINETIC ENERGY AND GENERALIZED FORCES 

11.6. Derive equations (5) and (6), page 283, for the work done on a system of particles. 

Suppose that a system undergoes increments dq lt dq 2 , . . . , dq n of the generalized coordinates. 
Then the rth particle undergoes a displacement 



" dr v 
dr v = 2t- dq a 
a=l d 9« 



Thus the total work done is 



dW = 2 F„«dr v = 2 1 2 F,« — W?„ = 2 * a dq a 
v=l v=l [a=l oq a j a =X 

N dr 

where * a — 2t F „ * t— 

v=i oq a 

We call * a the generalized force associated with the generalized coordinate q a . 



11.7. Prove that <i> a = dW/dq a . 

We have dW = ^j^-dq a . Also, by Problem 11.6, dW = 2*«<*g a . Then 

5 (♦.-£)*. = » 

Thus since the dq a are independent, all coefficients of dq a must be zero, so that * a = dW/dq a . 



11.8. Let F„ be the net external force acting on the vth particle of a system. Prove that 

d f^ . dr„~| ^ _ . dr v "V i? dTv 

7u \Z, mvrv • - — y - > m v r„ • t— = 2/ F " * ^r 
dt \v dq a ) v 5< Z« v oQct 

By Newton's second law applied to the yth particle, we have 

m„r„ = F„ (1) 



288 LAGRANGE'S EQUATIONS [CHAP. 11 

dr v dr„ 

Then m v r v >-— = F„ • — - (2) 

dq tt dq a 

d /. 3r„\ .. dr v . d / $ r v 

Now by Problem 11.4, -rr ( r„ • - — ) = r„ • - — + r„ • — [ - — 

dt\ v dq a J " dq a v dt\dq a 

_ v . ^ + ; . d ' v <* 

dq a dq a 

_,, .. 9t v d (• to v \ • &v m 

dq a dt\ dq a / Bq a 

Hence from (2) we have, since ra„ is constant, 

d ( . dr„\ . 3r„ dr v 

-r: ( ra„ r„ • - — 1 — m v r„ • - — = F„ • - — 
dt\ v " dq a J v dq a dq a 

Summing both sides with respect to v over all particles, we have 

d f« . 9r v \ v . 9r v « _ dr v 

d-t\^ m ^'^q-J-^ m ^-Iq- a = ? F ^'^ 

11.9. Let T be the kinetic energy of a system of particles. Prove that 
. . dT ^ . dic v ... dT ^ . dt v 

v ' Bq a v dq a w dQa „ dq a 

(a) The kinetic energy is T = r2%rj = - 2 m v K ' K- Thus 

dT ^ . 9r v 

— - 2,m v r v ' — 
dq a „ dq a 

(b) We have by the "cancellation of the dots" [Problem 11.3, page 286], 

dT v? • d *v V 3r " 

3g a " dq a dq a 



dr v 



LAGRANGE'S EQUATIONS 

11.10. Prove that ^ (|£) - J^ = *«, « = 1, . . .,n where *«=|F„ ^ . 
From Problem 11.8, 

d f« • 3r *l m • a *" - <? it . ar " 



U) 



From Problems 11.9(a) and 11.9(6), 

¥ m » r »'3T a ~ W« {S) 

Then substituting (2) and (3) in (1), we find 

d(8T_)_f_ = „ a W 

dt\dq a J dq a 



(5) 



The quantity Pa ~ J^~ 

is called the generalized momentum or conjugate momentum associated with the generalized coordi 
nate q a . 



CHAP. 11] 



LAGRANGE'S EQUATIONS 



289 



11.11. Suppose that the forces acting on a system of particles are derivable from a potential 
function V, i.e. suppose that the system is conservative. Prove that if L-T-V is 
the Lagrangian function, then 

dt\Bq a ) dq a 







If the forces are derivable from a potential V, then [see Problem 11.7], 

_ dW = dV 
*« 3q a Bq a 

Since the potential, or potential energy is a function of only the q's [and possibly the time t], 



«L = 4-<r-v) = ^?- 

Bq a dq a 3q a 



Then from Problem 11.10, 



d /BL\ _ BT_ _ _ 3V_ 
dt \9q a ) 9q a dq a 



d /3L\ 
dt\9qj 



dL 



= 



11.12. (a) Set up the Lagrangian for a simple pendulum and 
(b) obtain an equation describing its motion. 

(a) Choose as generalized coordinate the angle e made by string 
OB of the pendulum and the vertical OA [see Fig. 11-4]. 
If I is the length of OB, then the kinetic energy is 

T = £mv 2 = %m{le)2 = £mP« 2 (1) 

where m is the mass of the bob. 

The potential energy of mass m [taking as reference 
level a horizontal plane through the lowest point A] is given 

by 

V = mg(OA - OC) = mg(l — I cos e) 



= mgl(l — cos e) 



(«) 




Thus the Lagrangian is 

(b) Lagrange's equation is 
From (3), 



L = T 



dL 
Be 



V = ^ml 2 e 2 - mgl(l — cos e) 



dt\Bi) 
■ — mgl sin 9 , 



dL 
Be 



= 
3L 



Be 



= ml 2 e 



Substituting these in (4), we find 

ml 2 % e + mgl sin e = 



e + -jrsin* = 



(4) 
(5) 



which is the required equation of motion [compare Problem 4.23, page 102]. 



11.13. A mass Af 2 hangs at one end of a string which passes over a fixed frictionless non- 
rotating pulley [see Fig. 11-5 below]. At the other end of this string there is a 
non-rotating pulley of mass Mi over which there is a string carrying masses mi and m%. 
(a) Set up the Lagrangian of the system, (b) Find the acceleration of mass Af«. 

Let X t and X 2 be the distances of masses M x and M 2 respectively below the center of the fixed 
pulley. Let x x and x 2 be the distances of masses m x and m 2 respectively below the center of the 
movable pulley M v 

Since the strings are fixed in length, 

X t + X 2 = constant = a, x x + x 2 = constant = 6 



290 



LAGRANGE'S EQUATIONS 



[CHAP. 11 




Then by differentiating with respect to time t, 
X x + X 2 = or X 2 = —X x 
and x x + x 2 = or x 2 = —x x 

Thus we have 

Velocity of M x = X x 

Velocity of M 2 = X 2 = -X x 

d • • 

Velocity of m x — -j- (X x + x x ) = X x + x x 

Velocity of m 2 = — (X x + x 2 ) = X x + x 2 = X x — x x 

Then the total kinetic energy of the system is Flg * 11_5 

T = ±M X X 2 X + \M 2 ±\ + ±m x (X x + x x )2 + Lm 2 (k x -x x )* (1) 

The total potential energy of the system measured from a horizontal plane through the center 
of the fixed pulley as reference is 

V = -M 1 gX l - M 2 gX 2 - m x g(X x + x x ) - m 2 g{X x + x 2 ) 

= -M x gX x - M 2 fif(a-Z]) - m 1 g(X 1 + x 1 ) - m 2 g(X 1 + b - x x ) (2) 

Then the Lagrangian is 

L = T - V 

= %M X X.\ + \M 2 k\ + %m x (X x + x x )2 + \m 2 {X x - xtf 

+ M x gX x + M 2 g{a-X x ) + m 1 g(X x + x 1 ) + m 2 g(X x + b - x x ) (3) 



Lagrange's equations corresponding to X x and x x are 

Xj 3*1 






_d fdL_\ _dL_ _ 
dt\dx x J dx x 



U) 



From (3) we have 
jg- = M x g - M 2 g + m x g + m 2 g 



(M x - M 2 + m x + m 2 )g 



dL •••#•• • • 

—r- — M X X X + M 2 X X + m x (X x + x x ) + m 2 (X x — x x ) = (M x + M 2 + m x + m 2 )X x + (m x — m 2 )x x 
dX x 

dL 

— = m x g - m 2fl r = {m x -m 2 )g 

dL • • • . • . 

— — = mi(X x + x x ) — m 2 {X x — x x ) = (m x — m 2 )X x + (m x + m 2 )x x 
dXi 

Thus equations (4) become 

(M x + M 2 + m x + m 2 )X x + (m x — m 2 ) x x = (M x — M 2 + m x + m 2 )g 
(m x — m 2 )X x + (m x + m^) x x = (m x — m 2 )g 



Solving simultaneously, we find 



X, 



x x 



(M x — M 2 )(m x + m 2 ) + 4m t m 2 
(M x + M 2 )(m x + m 2 ) + 4m 1 m 2 

2M 2 (m x — m 2 ) 
(M x + M 2 )(m x + m 2 ) + 4m x m 2 



Then the downward acceleration of mass M 2 is constant and equal to 

(M 2 — M x )(m x + m 2 ) — 4m x m 2 



X 2 — —X x — 



(M x + Af 2 )(m 1 + m 2 ) + 4m 1 ra 2 



CHAP. Ill 



LAGRANGE'S EQUATIONS 



291 



11.14. Use Lagrange's equations to set up the differential equation of the vibrating masses 
of Problem 8.1, page 197. 

Refer to Figs. 8-7 and 8-8 of page 197. The kinetic energy of the system is 

T = \mx\ + \mx\ CO 

Since the stretches of springs AP,PQ and QB of Fig. 8-8 are numerically equal to x lf x 2 - x x and 
x 2 respectively, the potential energy of the system is 

V = %kx 2 1 + %k(x 2 -x 1 ) 2 + &x 2 2 (*) 

Thus the Lagrangian is 

L = t - V = \mx\ + \mx\ - \kx\ - \k[x 2 - xj* - &x\ 



Lagrange's equations are 



d /BL\ SL _ d /dL\ 

dt{axj dx i ' dt \dxJ 



dL 

dx 9 



= 



Then since 



dL 



dL 



= -kx, + k(x 2 - *i) = "(a?2 - 2x i)> T*~ - mx i 

dx-, dx l 



dL 

dx 2 



-k{x 2 — * x ) — kx 2 = «(«! — 2x 2 ), 



dL 
dX 2 



mx 2 



equations (4) become m'x\ = k(x 2 -2xJ, mx 2 - k(x 1 -2x 2 ) 

agreeing with those obtained in Problem 8.1, page 197. 



(3) 



(-4) 



(5) 



11.15. Use Lagrange's equations to find the differential equation for a compound pendulum 
which oscillates in a vertical plane about a fixed horizontal axis. 

Let the plane of oscillation be represented by the xy plane 
of Fig. 11-6, where O is its intersection with the axis of rota- 
tion and C is the center of mass. 

Suppose that the mass of the pendulum is M, its moment 
of inertia about the axis of rotation is I = MK 2 [K = radius 
of gyration], and distance OC = h. 

If e is the instantaneous angle which OC makes with^the 
vertical axis through O, then the kinetic energy is T = %I 6 2 = 
±MK 2 2 . The potential energy relative to a horizontal plane 
through is V = —Mgh cos 6. Then the Lagrangian is 
L = T -V = \MK 2 e 2 + Mgh cos e 

Since dL/do - -Mgh sin e and dLIdd - MK 2 e, Lagrange's 
equation is 



d_ / dL 
dt[do 



dL 
de 



i.e., 



MK 2 6 + Mgh sin = or 
Compare Problem 9.24, page 237. 



V + ^ sin « = 




Fig. 11-6 



11.16. A particle of mass m moves in a conservative force field. Find (a) the Lagrangian 
function, (b) the equations of motion in cylindrical coordinates ( P , 4>, z) [see Problem 
1.147, page 32]. 

(a) The total kinetic energy T - %m[p 2 + p 2 $ 2 + z 2 } . The potential energy V = V(p,<p,z). Then 
the Lagrangian function is 

L = T-V = lm[p2 + p 2^2 + |2] -V( p ,0,s) 



292 LAGRANGE'S EQUATIONS [CHAP. 11 

(b) Lagrange's equations are 

d /8L\ dL _ . d. • / .„ dV\ .. . dV 

d /8L\ dL _ . & 2 \,dV n d , „v dV 

d fdL\ dL n . d..,av n .. d v 



11.17. Work Problem 11.16 if the particle moves in the xy plane and if the potential 
depends only on the distance from the origin. 

In this case V depends only on P and z = 0. Then Lagrange's equations in part (6) of Problem 
11.16 become 

rnG-p'^) = -^, A( P 20) = o 
These are the equations for motion in a central force field obtained in Problem 5.3, page 122. 



LAGRANGE'S EQUATIONS FOR NON-HOLONOMIC SYSTEMS 

11.18. Derive Lagrange's equations (15), page 284, for non-holonomic constraints. 

Assume that there are m constraint conditions of the form 

2 A a dq a + Adt = 0, 2 B a dq a + B dt = 0, ... (1) 

a a 

where m < n, the number of coordinates q a . 
As in Problem 11.10, page 288, we have 

v _ d / dT \ 8T x .. dr v 

If Sr„ are virtual displacements which satisfy the instantaneous constraints [obtained by consider- 
ing that time t is a constant], then 

^ dr v 
Sr„ = 2^-5g a (3) 

Now the virtual work done is 

dr 
SW = 2 «**¥„ • 8r v = 2 2 m„ r„ • — - S? a = 2 Y a 8q a (4) 

v v a oq a a 

Now since the virtual work can be written in terms of the generalized forces <t> a as 

SW = 2* a S 9a (5) 

a 

we have by subtraction of (4-) and (5), 

2(^a-*a)Sg a = (6) 

a 

Since the 8q a are not all independent, we cannot conclude that Y a = $ a which would lead to 
Lagrange's equations as obtained in Problem 11.10. 

Prom (1), since t is constant for instantaneous con train ts, we have the m equations 

J,A a 8q a = 0, lB a Sq a = 0, ... (7) 

a a 

Multiplying these by the m Lagrange multipliers X lt X 2 , ... and adding, we have 

2 (M a + X 2 B a + • • •) 8q a = (8) 



CHAP. 11] 



LAGRANGE'S EQUATIONS 



293 



Subtraction of (6) and (8) yields 

2 (Y a - *« - M B - X 2 B a ) 8q a = 



(9) 



Now because of equations (7) we can solve for m of the quantities 8q a [say 5^, . . .,8q m ] in terms 
of the remaining 8q a [say 8g m + 1 , • • ., 8q n ]. Thus in (9) we can consider 8 ffl> . . ., S<? m as dependent 
and Sg m + i, . . ., 8q n as independent. 

Let us arbitrarily set the coefficients of the dependent variables equal to zero, i.e., 

(10) 



*a — X l^a ~ X 2 B a 



= o, 



1,2, ...,m 



Then there will be left in the sum (9) only the independent quantities 8q a and since these are ar- 
bitrary it follows that their coefficients will be zero. Thus 



Equations (2), (10) and (11) thus lead to 



= 0, 



m+1, . . .,« 



(«) 



d/3T\_|T = * a+Xl A a + X a F a + ••■ « = l,2,...,n 

as required. These equations together with (i) lead to n + m equations in n + m unknowns. 



11.19. Derive equations (16), page 284, for conservative non-holonomic systems. 

From Problem 11.18, 



d /dT 



dT 



= <f> a + X x A a + X 2 B« + 



(1) 



dt \dq a J d <la 

Then if the forces are derivable from a potential, <t>« = -dV/dq a where V does not depend on 
q a . Thus (1) can be written 



where L = T — V. 



dt\dq a J d <la 



11.20. A particle of mass m moves under the influence 
of gravity on the inner surface of the paraboloid 
of revolution x 2 + y 2 = az which is assumed 
frictionless [see Fig. 11-7]. Obtain the equations 
of motion. 

By Problem 11.16, the Lagrangian in cylindrical co- 
ordinates is given by 



L = ±m('p 2 + P 2 2 + I 2 ) - mgz 



{D 



Since x 2 + y 2 = p 2 , the constraint condition is p 2 — az = 
so that 



2p 8p — a 8z 







(2) 



If we call Qi = p, q% = <t>, <lz — z and compare (2) with 
the equations (7) of Problem 11.18, we see that 




Fig. 11-7 



A x = 2p, A 2 = 0, A 3 = -a (3) 

since only one constraint is given. Lagrange's equations [see Problem 11.19] can thus be written 



d /1L\ -*k - xA 



1,e " <tt\dp) 3p 

Using (1), these become 



2X lP , 



a = 1,2, 3 



<ft^4 



d^ 



o, 



dt\ dz 



dz 



XjO. 



294 LAGRANGE'S EQUATIONS [CHAP. 11 

m('p-p$ 2 ) = 2X lP (4) 

m'z = —mg — \ t a (6) 

We also have the constraint condition 

2pp — az = (7) 

The four equations &), (5), (6) and (7) enable us to find the four unknowns p, $, z, \ v 

11.21. (a) Prove that the particle of Problem 11.20 will describe a horizontal circle in the 
plane z = h provided that it is given an angular velocity whose magnitude is 
(o = y/2g/a . 

(b) Prove that if the particle is displaced slightly from this circular path it will 
undergo oscillations about the path with frequency given by (lh)^2g/a . 

(c) Discuss the stability of the particle in the circular path. 

(a) The radius of the circle obtained as the intersection of the plane z = h with the paraboloid 

p 2 = az is . — 

Po = Va>h (1) 

Letting z = h in equation (6) of Problem 11.20, we find 

Xi = —mg/a {2) 

Then using (1) and (*) in equation (4) of Problem 11.20 and calling = w, we find m(- Po u 2 ) = 

2{—mg/a)p Q or w 2 = 2g/a from which 

<o = V2g/a (3) 

The period and frequency of the particle in this circular path are given respectively by 

P 1 = 2,^ and /,=£>/? «» 

(b) From equation (5) of Problem 11.20, we find 

p 2 = constant = A (5) 

Assuming that the particle starts with angular speed w, we find A = aha so that 

^ = ahu/p 2 (6) 

Since the vibration takes place very nearly in the plane z — h, we find by letting z = h 
in equation (6) of Problem 11.20 that 

\j = —mg/a (7) 

Using (6) and (7) in equation (4) of Problem 11.20, we find 

P - a 2 h 2 a 2 /p 3 = -2£p/a (8) 

Now if the path departs slightly from the circle, then p will depart slightly from p . Thus we 
are led to make the transformation 

p = po + u (9) 

in (8), where u is small compared with p . Then (8) becomes 



(p + u) 6 a 



But to a high degree of approximation, 



1-1 _ i/ 1+ iiV 3 = l/i_ 3w 



(p + tt) 3 p3(l+w/p )3 p3\ p o y p 3\ p 

by the binomial theorem, where we have neglected terms involving u 2 , u s , . . . . Using the 
values of p and a given by (1) and (5) respectively, (10) becomes 



CHAP. 11] LAGRANGE'S EQUATIONS 295 

u 4- (Sg/a)u = ( 5 ) 

whose solution is it = e t cos V8«r/o t + e 2 sin VSg/a t. Thus 

p = Po + u = Vah + ej cos VSgJa t + e 2 sin y/Sg/a t 

It follows that if the particle is displaced slightly from the circular path of radius Po = Vah, 
it will undergo oscillations about the path with frequency 

t> - \~°L (7) 

or period 2 ~ v "\2g 

It is interesting that the period of oscillation in the circular path given by (4) is twice 
the period of oscillation about the circular path given by (7). 

(c) Since the particle tends to return to the circular path when it is displaced slightly from it, 
the motion is one of stability. 

11.22. Discuss the physical significance of the Lagrange multipliers Xi, A 2 , . . . in Problem 
11.18. 

In case there are no constraints the equations of motion are by Problem 11.10, 

±/dT\ _ 9T_ = 
dt \dq a ) d Qa 

In case there are constraints the equations are by Problem 11.18, 

dt \dq a J d< *« 
It follows that the terms \ x A a + \ 2 B a + • • • correspond to the generalized forces associated with 
constraints. 

Physically, the Lagrange multipliers are associated with the constraint forces acting on the 
system. Thus when we determine the Lagrange multipliers we are essentially taking into account 
the effect of the constraint forces without actually finding these forces explicitly. 



LAGRANGE'S EQUATIONS WITH IMPULSIVE FORCES 
11.23. Derive the equations (20), page 285. 

For the case where forces are finite we have by Problem 11.10, 

dt{ dq J dq a 
where *« = 2 F„ • t— ( g ) 

v °*ia 

Integrating both sides of (1) with respect to t from t = to t = r, 

Jo dt \*Qa) Jo a< *« J 



(5) 



so that ( —7- ) — 

\dq a /t=T 



Taking the limit as r -*■ 0, we have 
lim 



-lim f*Ldt = 2 (Aim C *,*)>£ 



auSL-(SL>-»rs* - n*s;'-*)'Z 



296 



LAGRANGE'S EQUATIONS 



[CHAP. 11 



f) -(f) = JA^ = T. 
a — dt = since - — is finite, and lim I F v dt = tf„. 

«i °tfa »9o t-»0 J„ 




Fig. 11-8 



11.24. A square ABCD formed by four rods of length 
21 and mass m hinged at their ends, rests on a 
horizontal frictionless table. An impulse of 
magnitude 3 is applied to the vertex A in the 
direction AD. Find the equations of motion. 
After the square is struck, its shape will in gen- 
eral be a rhombus [Fig. 11-8]. 

Suppose that at any time t the angles made by 
sides AD (or BC) and AB (or CD) with the x axis 
are e t and 6 2 respectively, while the coordinates of 
the center M are (x, y). Thus x,y,0 x ,6 2 are the gen- 
eralized coordinates. 

From Fig. 11-8 we see that the position vectors 
of the centers E,F,G,H of the rods are given re- 
spectively by 

r E = (x — I cos 0{)i + (y — I sin e t )j 
r F = (x + I cos 6 2 )\ + (y — I sin e 2 )j 
r G = (x + I cos 0i)i + (y + I sin e^j 
r H = {x — I cos e 2 )i + (y + I sin 6 2 )j 
The velocities of E, F, G and H at any time are given by 

v E = r E = (x + I sin X eji + (y — I cos $i x )j 
v F = r F = (x — I sin e 2 $ 2 )i + (y — I cos o 2 e 2 )j 
v g = *g — {x — I sin $! #i)i + (y + I cos e x b i)j 
v H = r H = (x + I sin 2 #2)* + (if + ^ co . s *2 *2)j 

The kinetic energy of a rod such as AZ? is the same as the kinetic energy of a particle of 
mass m located at its center of mass E plus the kinetic energy of rotation about an axis through 
E perpendicular to the xy plane. Since the angular velocity has magnitude e 2 aR d the moment of 
inertia of a rod of length 21 about its center of mass is I AB = $ml 2 , the total energy of rod AB is 

Similarly, the total kinetic energies of rods BC, CD and AD are 

T BC = %mrj. + lI BC el, T CD = %mr% + $I CD »l, T AD = %mi 2 H + %I AD o\ 

Thus the total kinetic energy is [using the fact that I = I AB = I BC = Icd = £ m * 2 ] 
T — T AB + T BC + T CD + T AD 

= |m(r| + r F + r% + r 2 H ) + I(e\ + «f ) 

= im(4^2 + 4y 2 + 2Z 2 * 2 + 2J2* 2 .) + £ mZ 2(^ + ^ 

= 2m(x 2 + y 2 ) + %ml 2 {e\ + e 2 2 ) 

Let us assume that initially the rhombus is a square at rest with its sides parallel to the 
coordinate axes and its center located at the origin. Then we have 

x = 0, y = 0, <?! = v/2, e 2 = 0, x = 0, y = 0, ^ = 0, e 2 = 

If we use the notation ( H and ( ) 2 to denote quantities before and after the impulse is applied, we 
have „. 

'2Z\ = (4«i)i = ( " ) = (4my)i = 

Bx/x \ d V/i 









CHAP. 11] LAGRANGE'S EQUATIONS 297 

dT \ / ST \ 

— r ) = (Amx) 2 = Am'x ( — r ) = (Amy) 2 = 4m£ 

3*/ 2 V%/ 2 



(2) 

(S) 
04) 

(5) 



\™l/2 


/dT 

\de 2 


• ) = §raZ 2 2 = \mV-h 

/2 


(5).-®. = '■ 


or 


4m* = f x 


(fe) 2 "(fe)i = ftf 


or 


Amy - f y 


(i?),"(®i = ^ 


or 


JmPff! = y 6i 


{'df 2 ) 2 ~\dfj 1 = Te * 


or 


fmPffg = fe 2 


where for simplicity we have now removed the subscript ( ) 2 . 


To find f x , f y , f $i , f h we note that 






T* = 


2j„- 

V 





where ^ are the impulsive forces. We thus have 



dr A dr B dr r dr n 

dr A 3r B dr c dr D 

T. = Sa-£+ A-^ + A-^+ A, •■£- (T) 

3r A dr R 3r c 3r D 

3r A 3r B 9r c dr D 

Now from Fig. 11-8 we find the position vectors of A,B,C,D given by 

r A = (x — I cos 9 1 — I cos 2 )i + (2/ ~ I sin x + I sin 2 )j 
r B = (x — I cos tf ! + I cos 2 )i + (2/ — J sin e 1 — I sin fl 2 )j 
r c = (x + I cos x + I cos e 2 )i + (y + I sin 8 X — I sin $ 2 )j 
td = (x + I cos e x — I cos e 2 )i + (y + I sm$i + I sin 2 )J 

Since the impulsive force at A is initially in the direction of the positive y axis, we have . 

c£a = Si do) 

Thus equations (6)-(9) yield 

fx = 0, 7^ = J, y 9l = - J* cos e x , Te 2 = SI cos 6 2 (11) 

Then equations (1)-(U) become 

4m* = 0, Amy = J, fm/ 2 ^ = — SI cos tf^ |mZ 2 ff 2 = £1 cos 2 (-Z2) 



11.25. Prove that the kinetic energy developed immediately after application of the impulsive 
forces in Problem 11.24 is T = S 2 /2m. 

From equations (12) of Problem 11.24, we have 

s %s * %s 

* = °' V = 4m> * = ~8mT C0S ^' '* = 8m7 cos '* 



298 



LAGRANGE'S EQUATIONS 



[CHAP. 11 



Substituting these values in the kinetic energy obtained in Problem 11.24, we find 

<72 3 (12 

But immediately after application of the impulsive forces, e x = tt/2 and 6 2 = approximately. 
Thus (1) becomes T = J 2 /2m. 



MISCELLANEOUS PROBLEMS 

11.26. In Fig. 11-9, AB is a straight frictionless wire 
fixed at point A on a vertical axis OA such that 
AB rotates about OA with constant angular 
velocity <■>. A bead of mass m is constrained to 
move on the wire, (a) Set up the Lagrangian. 
(b) Write Lagrange's equations, (c) Determine 
the motion at any time. 

(a) Let r be the distance of the bead from point A of 
the wire at time *. The rectangular coordinates 
of the bead are then given by 

x = r sin a cos at 

y = r sin a sin at 

z — h — r cos a 

where it is assumed that at t — the wire is in 
the xz plane and that the distance from O to A is h. 

The kinetic energy of the bead is 

T = ±m(x 2 + y 2 + z 2 ) 

— ±m{(r sin a cos at — ar sin a sin at) 2 




Fig. 11-9 



+ (r sin a sin at + ar sin a cos at) 2 + (— r cos a) 2 } 



= ^m(r 2 + co 2 r 2 sin 2 a) 
The potential energy, taking the xy plane as reference level, is V = mgz = mgr(/i - r cos a). 
Then the Lagrangian is 

L = T — V — \m(r 2 + a 2 r 2 sin 2 a) - mg(h — r cos a) 



(b) We have 



flL 

— = m« 2 r sin 2 a + mg cos a, 

dr 



§L_ 
dr 



= mr 



and Lagrange's equation is ^ ( ^ 



dL 
dr 



— or 



i.e., 



mr — (ma 2 r sin 2 a + w# cos a) — 
r — (w 2 sin 2 a)r = fif cos a 



CO 



(c) The general solution of equation (1) with the right hand side replaced by zero is 

c e (a sin a)t -f C2 e- (co sin a)t 

Since the right hand side of (1) is a constant, a particular solution is -^-^-. Inus tne 



<o z sin z a 



general solution of (1) is 



fif cos a 



r = ^CasinaOt + C2 e-Ca, sin aK _ __ — 



to 2 sin z a 



This result can also be written in terms of hyperbolic functions as 



r = c 3 cosh (« sin a)i + «4 sinh (« sin a)t 



g cos a 
to 2 sin 2 a 



(2) 



(3) 



CHAP. 11] LAGRANGE'S EQUATIONS 299 

11.27. Suppose that in Problem 11.26 the bead starts from rest at A. How long will it take 
to reach the end B of the wire assuming that the length of the wire is It 

Since the bead starts from rest at t = 0, we have r = 0, r = at t = 0. Then from equa- 
tion (2) of Problem 11.26, 

, g cos a , n 

Ci + c 2 = o ■ o and c x — c 2 = 

1 * w 2 sin 2 a 

Thus c x = Co = -^ — t-tt- and (2) of Problem 11.26 becomes 

1 2w 2 sm 2 a v ' 

„ _ _JL£2!L£_ / e (<a sin a)t J- e -(u sin a)t\ — & C0S a /*\ 



2« 2 sin 2 a 



flr COS a , , , . . .... 

■4 — r-^— {cosh (w sm a)£ — 1} 



w- 4 Sir a 

which can also be obtained from equation (3) of Problem 11.26. When r = Z, (2) yields 

Z« 2 sin 2 a 



so that the required time is 



cosh (« sin a)* = 14- 

9 cos a 



* = -4-coshVl-f Z " 2si f a N ) 
w sm a \ flr cos 2 a / 

= dn^ ln {( 1+ ^^UA/fl + ^^) -1 



g COS 2 a / V I gr cos 2 a 



11.28. A double pendulum [see Problem 11.1(c) and Fig. 11-3, page 285] vibrates in a 
vertical plane, (a) Write the Lagrangian of the system, (b) Obtain equations for 
the motion. 

(a) The transformation equations given in Problem 11.2, page 286, 

x x = l x cos 9 X y x = l x sin $ x 

x 2 = l x cos X 4- Z 2 cos 6 2 V2 — h sin *i 4- l 2 sin e 2 

yield »! = -l x e x sin ^ ^ = l x e x cos ^ 

»2 = ~h 9 \ sin tf x — Z 2 2 sin e 2 y 2 = Z^ cos e x 4- l 2 e 2 cos e 2 

The kinetic energy of the system is 

T = $m x {x\ + y 2 x ) + ^m&l + fy 

= %m x l\e\ 4- \m 2 \$o\ 4- Z^a + 2Z 1 Z 2 * 1 * 2 cos (^ - * 2 )] 

The potential energy of the system [taking as reference level a plane at distance l t + l 2 
below the point of suspension of Fig. 11-3] is 

V = m x g[l x 4- Z 2 - l x cos e x ] + m 2 g[l x 4- Z 2 - (Z x cos 0! + Z 2 cos 2 )] 
Then the Lagrangian is 
L = T - V 

= \™>xA<>\ + %™>2{l 2 j\ + |*| + 2Z 1 Z 2 m 2 cos (e x - e 2 )) (i) 

-m x g[l x + l 2 -l x cos $ x ] - m2g[l x + l 2 - (l x cos e x 4- Z 2 cos 6 2 )] 

(b) The Lagrange equations associated with e x and 2 are 



300 LAGRANGE'S EQUATIONS [CHAP. 11 

From (1) we find 

dL/de x — —m 2 l x l 2 e x e 2 sm ( e i ~ e z) ~ m idh sin e \ ~ m 20h sin e x 

dL/de x = ml\e x + m 2 l\e x + m 2 l x l 2 e 2 c °s (<?i — o 2 ) 

dL/de 2 = m 2 l x l 2 e x e 2 sin (*i — 2) ~ m*zQh sin 2 

dL/de 2 = m 2 l\e 2 + m 2 l x l 2 9 x cos {e x — e 2 ) 

Thus equations {2) become 

m x f x e\ + m 2 l\ e\ + m 2 l x l 2 'e 2 cos (*i ~ #2) ~ rn 2 l x l 2 e 2 (e x — e 2 ) sin (e x — 2 ) 
- -m 2 l x l 2 6 x 6 2 sin (e x — e 2 ) — m x gl x sin e x — m 2 gl x sin e x 

and w 2 ?2 * 2 + m 2 l x l 2 «i cos (<?! - e 2 ) - m 2 l x l 2 e\(e x - e 2 ) sin {e x - o 2 ) 

= m 2 l x l 2 x e 2 sin (<?! — 9 2 ) — m 2 gl 2 sin <9 2 
which reduce respectively to 

(m x + m 2 )l\ e\ + m 2 l x l 2 e 2 cos (<?! - e 2 ) + m 2 l x l 2 e\ sin (e x - e 2 ) = -{m x + m 2 )gl x sin e x (3) 

and m 2 Z 2 *0 2 + m 2 l x l 2 'e\ cos (0 x -6 2 ) - m 2 l x l 2 o\ sm(o x -e 2 ) = -m 2 gl 2 sine 2 (4) 

11.29. Write the equations of Problem 11.28 for the case m x = m 2 = m and h = h = I 

Letting m y = m 2 , l x = Z 2 in equations (3) and (4) of Problem 11.28 and simplifying, they 
can be written 

21 e\ + I e 2 cos (e x - e 2 ) + lei sin {e x - e 2 ) = -2g sin e x {1) 

I 'e x cos (e x - e 2 ) + I 'e 2 - le\ sin {e x - e 2 ) = -g sin e 2 (2) 

11.30. Obtain the equations of Problem 11.29 for the case where the oscillations are assumed 
to be small. 

Using the approximations sin e = 6, cos e = 1 and neglecting terms involving 6 2 e, the equa- 
tions (1) and (2) of Problem 11.29 become 

2l'e x + l'e 2 - -2ge x 
l'e\ + l'o 2 = —go 2 

11.31. Find the (a) normal frequencies and (b) normal modes corresponding to the small 
oscillations of the double pendulum. 

(a) Let e x = A x cos*t, * 2 = A 2 cos<ot [or A x e^, A 2 e^] in the equations of Problem 11.30. 

Then they can be written 

2(g-lo> 2 )A x - 1<* 2 A 2 = 0| 

-lo> 2 A x + (g-l<* 2 )A 2 = Oj CO 

In order for A x and A 2 to be different from zero, we must have the determinant of the coefficients 
equal to zero, i.e., 

2{g - Zo) 2 ) -Zco 2 

-hfl g ~ to 2 



= 



or Z 2 co 4 - 4tow 2 + 2g 2 = 0. Solving, we find 

4lg ± y/l6l 2 g 2 - M 2 9 2 (2±V^)9 



,.& = 



2l 2 I 

{2 + y/2)g 2 _ ( 2 - y[2)g 

A = — 1 — ' W2 - I 



CHAP. Ill LAGRANGE'S EQUATIONS 301 



Thus the normal frequencies are given by 

Wl 1 l(2 + V2)g _ tt2 _ i J (2-V2)g 

A = 2^ = 2^\ I and ft ~ tor ~ & \ I 



w 



(b) Substituting a 2 = a* = (2 + -y/2)g/l in equations (1) of Part (a) yields 

A 2 = -V2A t U) 

This corresponds to the normal mode in which the bobs are moving in opposite directions. 
Substituting u 2 = w 2 = (2 — \/2 )gll in equations (1) of Part (a) yields 

A 2 = V2A! (5) 

This corresponds to the normal mode in which the bobs are moving in the same directions. 

11.32. (a) Set up the Lagrangian for the motion of a symmetrical top [see Problem 10.25, 
page 268] and (b) obtain the equations of motion. 

(a) The kinetic energy in terms of the Euler angles [see Problem 10.24, page 268] is 

T = \{h»\ + 1 2 4 + h4) = %h& sin2 e + g2) + y^ cos e + fo (i) 

The potential energy is V = mgl cos 8 (2) 

as seen from Fig. 10-18, page 269, since distance OC = I and the height of the center of 
mass C above the xy plane is therefore I cos 8. Thus 

L = T - V = -J/^ 2 sin2 e + £2) + |/ 3 (0 cos e + J)2 _ mflr ; C os (5) 

(6) 3L/d0 = i^ 2 sin 8 cos + / 3 (<£ cos 8 + ^){—<t> sin 8) + mflfZ sin * 

dL/dd = J^ 

3L/30 = 

dL/d$ — Irf sin 2 8 + I s ($ cos + 4>) cos 

3L/d^ = 

dL/dj, — 7 3 (0 cos 8 + ^) 

Then Lagrange's equations are 

±/9L\_9L - n A/^.\_^ = n <L/i>LL\ -Ok - 

or I\0 ~ h4> 2 sin cos + / 3 (^ cos + \p)$ sin — mflrZ sin = (4) 

— [7 t sin 2 6 + J 3 (0 cos 8 + $) cos 0] = (5) 

j^[/ 3 (^cos0 + ^)] = (6) 

11.33. Use the results of Problem 11.32 to obtain agreement with the equation of 
(a) Problem 10.29(&), page 270, and (b) Problem 10.27(a), page 270. 

(a) From equations (5) and (6) of Problem 11.32 we obtain on integrating, 

Itf sin 2 8 + I 3 ($ cos 8 + ^) cos 8 = constant = K (1) 

<£ cos * + ^ = A (2) 

Using (2) in (1), we find I-Jj> sin 2 + 7 3 A cos = J£ 

(6) Using (2) in equation (4) of Problem 11.32, we find 

ij o — Ji<£ 2 sin 8 cos + / 3 A<£ sin 8 = mgrZ sin 8 



302 



LAGRANGE'S EQUATIONS 



[CHAP. 11 



11.34. Derive Euler's equations of motion for a rigid body by use of Lagrange's equations. 

The kinetic energy in terms of the Euler angles is [see Problem 10.24, page 268] 
T = £(J lW ? + I 2 4 + 7 8 „i) 

- -J/^ sin <? sin ,/< + 6 cos \J-) 2 + £/ 2 (<£ sin e cos ^ - 6 sin ^) 2 + %h(4> cos + «/<) 2 

Then dT/dxp - I X Q> sin sin ^ + (9 cos </<)(0 sin e cos^p — e sin ./>) 

+ I 2 (<£ si n * cos \p — 6 sin ^)(— <£ sin sin >/< — cos ^) 

= / 1 w 1 w 2 + J 2 (w 2 )(— a^) = (^1 — -^2) w l w 2 

ar/af = i 3 ($ cos + «/>) = -?3 w 3 

Then by Problem 11.10, page 288, Lagrange's equation corresponding to ^ is 

d_/dT\ 9T _ 

or ^3"3 + (h ~ ^i) w i"2 = *</; (■*) 

This is Euler's third equation of (22), page 256. The quantity *^ represents the general- 
ized force corresponding to a rotation ^ about an axis and physically represents the component 
A 3 of the torque about this axis [see Problem 11.102]. 

The remaining equations , . /T t ^ . ,„\ 

7 lWl + (I 3 - i 2 ) w 2 w 3 = A l \ 2 ) 

J 2 « 2 + (h - ^3)"3«l = A 2 W 

can be obtained from symmetry considerations by permutation of the indices. They are not directly 
obtained by using the Lagrange equations corresponding to and <£ but can indirectly be deduced 
from them [see Problem 11.79]. 



11.35. A bead slides without friction on a f rictionless 
wire in the shape of a cycloid [Fig. 11-10] 
with equations 

x = a(0 — sin 9), y = a(l + cos 9) 

where ^ 9 ^ 2tt. Find (a) the Lagrangian 
function, (b) the equation of motion. 

(a) Kinetic energy = T = -|m(x 2 + y 2 ) 

= lma 2 {[(l - cos e)d] 2 + [-sin e] 2 } 
— ma 2 (l — cos e)e 2 

Potential energy = V - mgy — mga{l + cos 0) 

Then 

Lagrangian = L = T -V = ma 2 (l - cos 0)0 2 - mga(l + cos 0) 




Fig. 11-10 



... d fdL\ dL n . 



d 



[2ma 2 (l — cos 0)0] — [ma 2 sin 6 e 2 + mga sin e] = 

9 



dt 
[(1 - cos 9)e] - \ sin 6 2 - — sin 6 = 



which can be written (1 - cos 6) e + % sin 6 e 2 - g- sin 6 - 



11.36. (a) Show that the equation of motion obtained in part (b) of Problem 11.35 can be 

written , 2 

a u + SL W = o where w = cos (5/2) 



dt 2 



4a 



CHAP. 11] LAGRANGE'S EQUATIONS 303 

and thus (b) show that the bead oscillates with period 2 7 r\ / 4a/g . 

(a) If u = cos (0/2), then 

~r = -± sin (0/2)0, ^ = -A sin (0/2)0* -A cos (0/2)02 
Thus -v-g- + j-u = is the same as 



\ sin (0/2) V - i cos (0/2)02 + ■■&- cos (0/2) = 

+ i cot (0/2)0*2 - |L cot (0/2) = (i) 



2 —--'-'- 4~~v,-/- ■ 4a 

which can be written as 



Since cot (0/2) = ^ os (<?/2) = 2 sin (0/2) cos (0/2) sin 

w ' sin (0/2) 2 sin2 (0/2) 1 - cos 

it follows that equation (1) is the same as that obtained in Problem 11.35(&). 

(b) The solution of the equation is 

u = cos (0/2) = Cj cos V4a/flr t + c 2 sin y/ba/g t 
from which we see that cos (0/2) returns to its original value after a time 2iry/£a/g which 
is the required period. Note that this period is the same as that of a simple pendulum with 
length I — 4a. 

An application of this is the cycloidal pendulum. See Problem 4.86, page 112. 

11.37. Obtain equations for the rolling sphere of Problem 9.42, page 244 by use of Lagrange's 
equations. 

Refer to Fig. 9-33 in which <t> and ^ represent generalized coordinates. Since the sphere of 
radius CP = a rolls without slipping on the sphere of radius OP = b, we have 

6 d<p/dt — a d\f//dt or b$ = af 
which shows that if <p = when ^ — 0, then 

b<j> = a\p (i) 

Thus </> and $ [and therefore d<p and d> or 8<t> and 3^] are not independent. 
The kinetic energy of the rolling sphere is 

T = im(a + 6)2^2 + _ij w 2 

= ±m(a + 6)2^2 + ^(|ma2 )( ; + ^2 

using the fact that I = %ma?- is the moment of inertia of the sphere about a horizontal axis 
through its center of mass. 

The potential energy of the rolling sphere [taking the horizontal plane through O as reference 
level] is 

V = mg(a + b) cos <f> 

Thus the Lagrangian is 

L = T - V = {m(o + 6) 2 02 + ±ma 2 (4> + J)2 - m g(a + b) cos <f> (2) 

We use Lagrange's equations (16), page 284, for non-holonomic systems. From (1) we have 

b 8<f> — a S^ = (#\ 

so that if we call g x = and q 2 - f and compare with equation (7) of Problem 11.18, page 292 
we find ' 

A t = b, A 2 = —a (4) 

Thus equations (16), page 284, become 

d L /dL\ dL _ 

_d /BL\ dL _ 



304 LAGRANGE'S EQUATIONS [CHAP. 11 

Substitution of (2) intd (5) and (6) yields 

m(a+b) 2 $ + fma 2 (0 + £) - mg(a + b) sin <f> - \ x b (7) 

fraa 2 (0 + 'vfr) = -X x a (8) 

Substituting ^ = (b/a)</> [from (1)] into (7) and (8), we find 

m(a + b) 2 5? + §raa 2 (l + 6/a) £ - wfir(o + 6) sin <p = \ x b (9) 

. . fma 2 (l + 6/a)0 = -XjO (10) 

Now from (10) we have Xi = —%m(a + b)$ 

and using this in (9) it becomes after simplifying and solving for *<p, 

•• 5flf 

* ~ 7(^+6) 8m * 

This is the same equation as that of (2) in Problem 9.42, page 244, with <f> - w/2 - e. To find the 
required angle at which the sphere falls off, see Problem 11.104. 

11.38. (a) Solve the equations of motion obtained in Problem 11.24, page 296, and (b) inter- 
pret physically. 

(a) From the first of equations (12) in Problem 11.24 we have 

<c = constant = (0 

since x = at t = 0. Similarly, from the second of equations (12) we have 

* = is' « 

since y — at t = 0. 

From the third of equations (12) we find on separating the variables, 

or on integrating, In cot I ^ - — J - ~g^J + c i 



i.e. 



tan(|-^) = c 2 eW* ml 



Thus since X = n-/2 at < = 0, we have c 2 = 0. This means that for all time we must 
have $i = n72. 

From the fourth of equations (12) in Problem 11.24 we have similarly, 

sec 9 2 de 2 - z^jdt 

e 2 \ 3Jt 



or on integrating, In cot (7 — — J - -^—f + c 3 

i.e., tan (i"^) =' W~ ZSmml 

Now when t = 0, e % = so that c 4 = 1. Then 

ten/E-M = e -8^8«i or , 2 = |- 2 tan-i( e - 3 ^ /8w 

(6) Equations (1) and (2) show that the center moves along the y axis with constant speed J/4m. 
The rods AD and BC are always parallel to the y axis while rods AB and CD slowly rotate until 
finally [*-»"] the rhombus collapses, so that all four rods will be on the y axis. 



CHAP. 11] LAGRANGE'S EQUATIONS 305 

Supplementary Problems 

GENERALIZED COORDINATES AND TRANSFORMATION EQUATIONS 

11.39. Give a set of generalized coordinates needed to completely specify the motion of each of the follow- 
ing: (a) a bead constrained to move on a circular wire; (6) a particle constrained to move on a 
sphere; (c) a compound pendulum [see page 228]; (d) an Atwood's machine [see Problem 3.22, 
page 76]; (e) a circular disk rolling on a horizontal plane; (/) a cone rolling on a horizontal plane. 

11.40. Write transformation equations for the motion of a triple pendulum in terms 'of a suitable set of 
generalized coordinates. 

11.41. A particle moves on the upper surface of a frictionless paraboloid of revolution whose equation 
is x 2 + y 2 = cz. Write transformation equations for the motion of the particle in terms of a suit- 
able set of generalized coordinates. 

11.42. Write transformation equations for the motion of a particle constrained to move on a sphere. 

CLASSIFICATION OF MECHANICAL SYSTEMS 

11.43. Classify each of the following according as they are (i) scleronomic or rheonomic, (ii) holonomic 
or non-holonomic, and (iii) conservative or non-conservative: 

(a) a horizontal cylinder of radius a rolling inside a perfectly rough hollow horizontal cylinder of 

radius b > a; 
(6) a cylinder rolling [and possibly sliding] down an inclined plane of angle a; 

(c) a sphere rolling down another sphere which is rolling with uniform speed along a horizontal 
plane; 

(d) a particle constrained to move along a line under the influence of a force which is inversely 
proportional to the square of its distance from a fixed point and a damping force proportional 
to the square of the instantaneous speed. 

Ans. (a) scleronomic, holonomic, conservative 

(b) scleronomic, non-holonomic, conservative 

(c) rheonomic, non-holonomic, conservative 

(d) scleronomic, holonomic, non-conservative 

WORK, KINETIC ENERGY AND GENERALIZED FORCES 

11.44. Prove that if the transformation equations are given by r v = r v (q v q 2 , . . ., q n ), i.e. do not involve 
the time t explicitly, then the kinetic energy can be written as 

n n 

T = 2 2 a a a la 9b 
where a a/3 are functions of the q a . 

11.45. Discuss Problem 11.44 in case the transformation equations depend explicitly on the time t. 

11.46. If F(Xx, Xy, Xz) = X n F(x, y, z) where X is a parameter, then F is said to be a homogeneous function 
of order n. Determine which (if any) of the following functions are homogeneous, giving the order 
in each case: 

(a) x 2 + 2/ 2 + z 2 + xy + yz + xz (e) x s tan -1 (y/x) 

(6) 3x-2y + 4z (/) 4 sin xy 

(c) xyz + 2xy + 2xz + 2yz (g) (x + y + z)/(x 2 + y 2 + z 2 ) 

(d) (x + y + z)/x 

Ans. (a) homogeneous of order 2, (6) homogeneous of order 1, (c) non-homogeneous, (d) homo- 
geneous of order zero, (e) homogeneous or order 3, (/) non-homogeneous, (g) homogeneous of 
order —1. 

11.47. If F{x,y,z) is homogeneous of order n [see Problem 11.46], prove that 

dF , dF , dF 

a— + y-z h 2t- = nF 

dx " by dz 

This is called Euler's theorem on homogeneous functions. 

[Hint. Differentiate both sides of the identity F(\x,\y,\z) - X n F(x,y,z) with respect to X and 

then place X = 1.] 

11.48. Generalize the result of Problem 11.47. 



306 LAGRANGE'S EQUATIONS [CHAP. 11 

11.49. Prove that if the transformation equations do not depend explicitly on time t, and T is the kinetic 
energy, then SJ , f 

Can- you prove this directly without the use of Euler's theorem on homogeneous functions [Problem 
11.47]? 

LAGRANGE'S EQUATIONS 

11.50. (a) Set up the Lagrangian for a one dimensional harmonic oscillator and (6) write Lagrange's 
equations. Ans. (a) L = ±mx 2 — ^kx 2 , (b) mx + KX = 

11.51. (a) Set up the Lagrangian for a particle of mass m falling freely in a uniform gravitational field 
and (b) write Lagrange's equations. 

11.52. Work Problem 11.51 in case the gravitational force field varies inversely as the square of the dis- 
tance from a fixed point O assuming that the particle moves in a straight line through 0. 

11.53. Use Lagrange's equations to describe the motion of a particle of mass m down a frictionless in- 
clined plane of angle a. 

11.54. Use Lagrange's equations to describe the motion of a projectile launched with speed v at angle 
a with the horizontal. 

11.55. Use Lagrange's equations to solve the problem of the (a) two-dimensional and (6) three- 
dimensional harmonic oscillator. 

11.56. A particle of mass m is connected to a fixed point P on a horizontal plane by a string of length 
I. The plane rotates with constant angular speed w about a vertical axis through a point O of the 
plane, where OP = a. (a) Set up the Lagrangian of the system. (6) Write the equations of motion 
of the particle. 

11.57. The rectangular coordinates (as, y, z) defining the position of a particle of mass tn moving in a force 
field having potential V are given in terms of spherical coordinates (r, 6, <f>) by the transformation 

equations 

x — r sin <f> cos e, y = r sin <£ sin e , z = r cos 

Use Lagrange's equations to set up the equations of motion. 
Ans. 



m[ 


r — r$ 2 — re 2 cos 2 0] = — — 




m 


~ d 
-tz (r 2 0) + r 2 e 2 sin <f> cos <t> 


= 


1 ay 

r Be 


m- 
c 


d , „. . „ x 1 d 
.. (r 2 e sin 2 </>) = — . . , 
it r sin e a 


V 





11.58. Work Problem 11.56 if the particle does not necessarily move in a straight line through O. 

11.59. Work Problem 4.23, page 102, by use of Lagrange's equations. 

LAGRANGE'S EQUATIONS FOR NON-HOLONOMIC SYSTEMS 

11.60. (a) Work Problem 11.20, page 293, if the paraboloid is replaced by the cone x 2 + y 2 = c 2 z 2 . 
(b) What modification must be made to Problem 11.21, page 294, in this case? 

11.61. Use the method of Lagrange's equations for non-holonomic systems to solve the problem of a 
particle of mass m sliding down a frictionless inclined plane of angle a. 

11.62. Work Problem 3.74, page 82 by using the method of Lagrange's equations for non-holonomic 
systems. 

LAGRANGE'S EQUATIONS WITH IMPULSIVE FORCES 

11.63. A uniform rod of length I and mass M is at rest on a horizontal frictionless table. An impulse of 
magnitude $ is applied to one end A of the rod and perpendicular to it. Prove that (a) the 
velocity given to end A is 4J/M, (b) the velocity of the center of mass is J/M and (c) the rod 
rotates about the center of mass with angular velocity of magnitude 6J/MI. 



CHAP. Ill 



LAGRANGE'S EQUATIONS 



307 



11.64. In Fig. 11-11, AB and BC represent two uniform rods 
having the same length I and mass M smoothly hinged 
at B and at rest on a horizontal frictionless plane. 
An impulse is applied at C normal to BC in the di- 
rection indicated in Fig. 11-11 so that the initial 
velocity of point C is v . Find (a) the initial vel- 
ocities of points A and B and (6) the magnitudes of 
the initial angular velocities of AB and BC about 
their centers of mass. 
Ans. (a) v /7, - 2v /7; (6) 3v /ll, - 9v /7l 



Fig. 11-11 



11.65. Prove that the total kinetic energy developed by the system of Problem 11.64 after the impulse 
is \Mv% 

11.66. A square of side a and mass M, formed from 4 uniform rods which are smoothly hinged at their 
edges, rests on a horizontal frictionless plane. An impulse is applied at a vertex in a direction 
of the diagonal through the vertex so that the vertex is given a velocity of magnitude v . Prove 
that the rods move about their centers of mass with angular speed 3v /4a. 

11.67. (a) If $ is the magnitude of the impulse applied to the vertex in Problem 11.66, prove that the 
kinetic energy developed by the rods is given by 5<5 2 /4ilf. (b) What is this kinetic energy in terms of 
i> ? (c) Does the direction of the impulse make any difference? Explain. 

11.68. In Problem 11.24, page 296, suppose that the impulse is applied at the center of one of the rods in 
a direction which is perpendicular to the rod. Prove that the kinetic energy developed is <jj 2 /8m. 



MISCELLANEOUS PROBLEMS 

11.69. A particle of mass m moves on the inside of a smooth hollow hemisphere of radius a having its 
vertex on a horizontal plane. With what horizontal speed must it be projected so that it will 
remain in a horizontal circle at height h above the vertex? 



11.70. A particle of mass m is constrained to move inside a 
thin hollow frictionless tube [see Fig. 11-12] which is 
rotating with constant angular velocity a in a hor- 
izontal xy plane about a fixed vertical axis through 
O. Using Lagrange's equations, describe the motion. 

11.71. Work Problem 11.70 if the xy plane is vertical. 

11.72. A particle of mass m moves in a central force field 
having potential V(r) where r is the distance from 
the force center. Using spherical coordinates, (a) 
set up the Lagrangian and (6) determine the equa- 
tions of motion. Can you deduce from these equa- 
tions that the motion takes place in a plane [compare 
Problem 5.1, page 121]? 




Fig. 11-12 



11.73. A particle moves on a frictionless horizontal wire of radius a, acted upon by a resisting force which 
is proportional to the instantaneous speed. If the particle is given an initial speed v , find the 
position of the particle at any time t. 

Ans. e = (?nv Ac)(l — e~ Kt/ma ) where e is the angle which a radius drawn to m makes with a fixed 
radius such that e = at t = 0, and k is the constant of proportionality. 



11.74. Work Problem 11.73 if the resisting force is proportional to the square of the instantaneous speed. 

no, \ m J 

11.75. A spherical pendulum is fixed at point O but is otherwise free to move in any direction. Write 
equations for its motion. 



11.76. Work Problem 9.29, page 239, by use of Lagrange's equations. 



308 



LAGRANGE'S EQUATIONS 



[CHAP. 11 



11.77. Work Problem 11.20 if the paraboloid of revolution is replaced by the elliptic paraboloid 
az = bx 2 + cy 2 where a, b, c are positive constants. 

11.78. Prove that the generalized force corresponding to the angle of rotation about an axis physically 
represents the component of the torque about this axis. 

11.79. (a) Obtain Lagrange's equations corresponding to B and <p in Problem 11.34, page 302, and show 
that these are not the same as equations (2) and (3) of that problem. (6) Show how to obtain 
equations (2) and (3) of Problem 11.34 from the Lagrange equations of (a). 



11.80. Two circular disks, of radius of gyrations K V K 2 and masses m u m 2 
respectively, are suspended vertically on a wire of negligible mass [see 
Fig. 11-13]. They are set into motion by twisting one or both of the 
disks in their planes and then releasing. Let X and e 2 be the angles 
made with some specified direction. 



V//////////////////A 



(a) Prove that the kinetic energy is 
T 



^(m.Kl'e 2 + m 2 K\el) 



(b) Prove that the potential energy is 

v = iMf + T^-*!) 2 ] 

where t x and t 2 are torsion constants, i.e. the torques required to 
rotate the disks through one radian. 

(c) Set up Lagrange's equations for the motion. 




Fig. 11-13 



11.81. Solve the vibrating system of Problem 11.80, finding (a) the normal frequencies and (6) the normal 
modes of vibration. 

11.82. Generalize the results of Problem 11.80 and 11.81 to 3 or more disks. 



11.83. (a) Prove that if m x ¥> m 2 and l x ¥= l 2 in the double pendulum of Problem 11.28, then the normal 
frequencies for small oscillations are given by w/2tt where 



(m t + m 2 )(h + h) ± V(»h + w 2 ) [«»!(«! - y 2 + m 2 (*i + l 2 ) 2 



2l 1 l 2 m 1 
(b) Discuss the normal modes corresponding to the frequencies in (a). 



9 



11.84. Examine the special case h - l 2 , m, ¥* m 2 in Problem 11.83. 

11.85. Use Lagrange's equations to describe the motion of a sphere of radius a rolling on the inner surface 
of a smooth hollow hemisphere of radius b > a. 

11.86. A particle on the inside surface of a frictionless paraboloid of revolution az = x 2 + y 2 at a height 
H x above the vertex is given a horizontal velocity v . Find the value of v in order that the particle 
oscillate between the planes z = H x and z = H 2 . Ans. v Q = y2gH 2 

11.87. Find the period of the oscillation in Problem 11.86. 



11.88. A sphere of radius a is given an initial velocity v up a frictionless inclined plane of angle a in a 
direction which is not along the line of greatest slope. Prove that its center describes a parabola. 

11.89. A bead of mass m is constrained to move on a frictionless horizontal circular wire of radius a 
which is rotating at constant angular speed a about a fixed vertical axis passing through a point 
on the wire. Prove that relative to the wire the bead oscillates like a simple pendulum. 



CHAP. Ill 



LAGRANGE'S EQUATIONS 



309 



11.90. If a particle of mass m and charge e moves with velocity v in an electric field E and magnetic 
field B, the force acting on it is given by 

F = e(E + v X B) 

In terms of a scalar potential * and a vector potential A the fields can be expressed by the relations 

E = -V* - dA/dt, B = V X A 

Prove that the Lagrangian defining the motion of such a particle is 

L — |«m; 2 + e(A • v) — e* 

11.91. Work Problem 10.86, page 278, by use of Lagrange's equations. 

11.92. A uniform rod of length I and mass M has its ends constrained to move on the circumference of a 
smooth vertical circular wire of radius a > 1/2 which rotates about a vertical diameter with con- 
stant angular speed u. Obtain equations for the motion of the rod. 



11.93. Suppose that the potential V depends on q v as well as q v . Prove that the quantity 

T + V 



is a constant. 



v . dV 
dq v 



11.94. Use Lagrange's equations to set up and solve the two body problem as discussed in Chapter 5 
[see for example page 121.] 



11.95. Find the acceleration of the 5 gm mass in the pulley system of 
Fig. 11-14. Ans. 71^/622 



11.96. 



A circular cylinder of radius a having radius of gyration K 
with respect to its center, moves down an inclined plane of 
angle a. If the coefficient of friction is n, use Lagrange's 
equations to prove that the cylinder will roll without slipping 
K 2 



if ii < 



tan a. Discuss the cases where n does not 



a 2 + K 2 
satisfy this inequality. 

11.97. Use Lagrange's equations to solve Problem 8.27, page 213. 

11.98. Describe the motion of the rods of Problem 11.64 at any time 
t after the impulse has been applied. 



r~\ 



15 gm 



10 gm 

I □ 2gm d 

M 7 « m 3gm 

5 gm 

Fig. 11-14 



11.99. In Fig. 11-15, AB represents a frictionless horizontal plane 

having a small opening at O. A string of length I which A_ 
passes through O has at its ends a particle P of mass m and 
a particle Q of equal mass which hangs freely. The particle 
P is given an initial velocity of magnitude v at right angles 
to string OP when the length OP = a. Let r be the in- 
stantaneous distance OP while e is the angle between OP and 
some fixed line through O. 

(a) Set up the Lagrangian of the system. 

(6) Write a differential equation for the motion of P in terms of r. 
(c) Find the speed of P at any position. 
Ans. (a) L = %m[2r 2 + r 2 e 2 ] + mg(l - r) 
(6) r = a 2/ i%/r 2 — g 



(c) r = y/2av? + 2g(a - r) - 2a 2 i%/r 



O 



IF 



Q 



Fig. 11-15 



11.100. Work Problem 11.99 if the masses of particles P and Q are ra x and m 2 respectively. 



310 



LAGRANGE'S EQUATIONS 



[CHAP. 11 



11.101. Prove that if v = yfag the particle P of Problem 11.99 remains in stable equilibrium in the circle 
r = a and that if it is slightly displaced from this e quilibrium position it oscillates about this 
position with simple harmonic motion of period 2irv2ct/3flr. 

11.102. Prove that the quantity fy in Problem 11.34, page 302, physically represents the component A 3 
of the torque. 

11.103. Describe the motion of the system of (a) Problem 11.63 and (6) Problem 11.66 at any time t after 
the impulse has been applied. 

11.104. Show how to find the angle at which the sphere of Problem 11.37, page 303, falls off. 

11.105. (a) Set up the Lagrangian for the triple pendulum of Fig. 11-16. 
(6) Find the equations of motion. 

11.106. Obtain the normal frequencies and normal modes for the triple pendulum of Problem 11.105 assuming 
small oscillations. 



11.107. Work Problems 11.105 and 11.106 for the case where the masses and lengths are unequal. 




m 



Fig. 11-16 



Fig. 11-17 



11.108. A vertical spring [Fig. 11-17] has constant k and mass M. If a mass m is placed on the spring and 
set into motion, use Lagrange's eq uations to prove that the system will move with simple harmonic 
motion of period 2v^(M + 3w)/3<c. 



Chapter 12 HAMILTON IAN 

THEORY 



HAMILTONIAN METHODS 

In Chapter 11 we investigated a formulation of mechanics due to Lagrange. In this 
chapter we investigate a formulation due to Hamilton known collectively as Hamiltonian 
methods or Hamiltonian theory. Although such theory can be used to solve specific prob- 
lems in mechanics, it develops that it is more useful in supplying fundamental postulates 
in such fields as quantum mechanics, statistical mechanics and celestial mechanics. 

THE HAMILTONIAN 

Just as the Lagrangian function, or briefly the Lagrangian, is fundamental to Chapter 
11, so the Hamiltonian function, or briefly the Hamiltonian, is fundamental to this chapter. 

The Hamiltonian, symbolized by H, is defined in terms of the Lagrangian L as 

n 

H = 2 Vaq a - L (1) 

a=l 

and must be expressed as a function of the generalized coordinates q a and generalized 
momenta p a . To accomplish this the generalized velocities q a must be eliminated from (1) 
by using Lagrange's equations [see Problem 12.3, for example]. In such case the function 
H can be written 

H(pi, .. .,p n , qi, ..., q n , t) (2) 

or briefly H(p a , q a , t), and is also called the Hamiltonian of the system. 

HAMILTON'S EQUATIONS 

In terms of the Hamiltonian, the equations of motion of the system can be written in 
the symmetrical form 

• dH 



dH 

qa — t — 

dp* 



(3) 



These are called Hamilton's canonical equations, or briefly Hamilton's equations. The 
equations serve to indicate that the p a and q a play similar roles in a general formulation 
of mechanical principles. 

THE HAMILTONIAN FOR CONSERVATIVE SYSTEMS 

If a system is conservative, the Hamiltonian H can be interpreted as the total energy 
(kinetic and potential) of the system, i.e., 

H = T + V (4) 

Often this provides an easy way for setting up the Hamiltonian of a system. 

311 



312 HAMILTONIAN THEORY [CHAP. 12 

IGNORABLE OR CYCLIC COORDINATES 

A coordinate q a which does not appear explicitly in the Lagrangian is called an ignorable 
or cyclic coordinate. In such case 

*■ = £ = » < 5 > 

so that p a is a constant, often called a constant of the motion. 
In such case we also have dH/dq a = 0. 



PHASE SPACE 

The Hamiltonian formulation provides an obvious symmetry between the p a and q a 
which we call momentum and position coordinates respectively. It is often useful to imagine 
a space of In dimensions in which a representative point is indicated by the In coordinates 

(Pu ■ • .,Pn, <7i, . . . , q n ) (6) 

Such a space is called a 2n dimensional phase space or a pq phase space. 

Whenever we know the state of a mechanical system at time t, i.e. we know all position 
and momentum coordinates, then this corresponds to a particular point in phase space. 
Conversely, a point in phase space specifies the state of the mechanical system. While the 
mechanical system moves in the physical 3 dimensional space, the representative point 
describes some path in the phase space in accordance with equations (3). 



LIOUVILLE'S THEOREM 

Let us consider a very large collection of conservative mechanical systems having the 
same Hamiltonian. In such case the Hamiltonian is the total energy and is constant, i.e., 

H(pi, ...,Pn,qi,...,Qn) = constant = E {7) 

which can be represented by a surface in phase space. 

Let us suppose that the total energies of all 
these systems lie between Ei and E 2 . Then the 
paths of all these systems in phase space will lie 
between the two surfaces H = Ei and H — E 2 
as indicated schematically in Fig. 12-1. 

Since the systems have different initial condi- 
tions, they will move along different paths in the 
phase space. Let us imagine that the initial points 
are contained in region %i of Fig. 12-1 and that 
after time t these points occupy region % 2 . For 
example, the representative point corresponding to 
one particular system moves from point A to 
point B. From the choice of %i and % 2 it is clear 
that the number of representative points in them 

are the same. What is not so obvious is the follow- Fig. 12-1 

ing theorem called Liouville's theorem. 

Theorem 12.1: Liouville's Theorem. The 2n dimensional volumes of %i and ^2 are the 
same, or if we define the number of points per unit volume as the density then the density 
is constant. 




CHAP. 12] 



HAMILTONIAN THEORY 313 



We can think of the points of %i as particles of an incompressible fluid which move from %i 
to %2 in time t. 

THE CALCULUS OF VARIATIONS 

A problem which often arises in mathematics is that of finding a curve y = Y(x) joining 
the points where x — a and x = b such that the integral 

>b 



f F(x,y,y')dx (8) 

*J a 



where y' - dy/dx, is a maximum or minimum, also called an extremum or extreme value. 
The curve itself is often called an extremal. It can be shown [see Problem 12.6] that a 
necessary condition for (8) to have an extremum is 

d_/dF\_ dF = (g) 

dx\dy'J by v ' 

which is often called Euler's equation. This and similar problems are considered in a 
branch of mathematics called the calculus of variations. 



HAMILTON'S PRINCIPLE 

The obvious similarity of (9) to Lagrange's equations leads one to consider the problem 
of determining the extremals of 



J»t 2 
L{Qu 



, Qn, Ql, . . ., On, t)dt (10) 

J-»t 2 
Ldt 
h 

where L = T — V is the Lagrangian of a system. 

We can show that a necessary condition for an extremal is 



dt\dq a J dq a 

which are precisely Lagrange's equations. The result led Hamilton to formulate a general 
variational principle known as 

Hamilton's Principle. A conservative mechanical system moves from time U to time 
t 2 in such a way that 

C*Ldt (12) 

sometimes called the action integral, has an extreme value. 

Because the extreme value of (12) is often a minimum, the principle is sometimes referred 
to as Hamilton's principle of least action. 

The fact that the integral (12) is an extremum is often symbolized by stating that 

8 Vhdt = (13) 

where S is the variation symbol. 



314 HAMILTONIAN THEORY [CHAP. 12 

CANONICAL OR CONTACT TRANSFORMATIONS 

The ease in solution of many problems in mechanics often hinges on the particular 
generalized coordinates used. Consequently it is desirable to examine transformations 
from one set of position and momentum coordinates to another. For example if we call 
q a and p a the old position and momentum coordinates while Q a and P a are the new position 
and momentum coordinates, the transformation is 

Pa = Pa (Pi, . . .,Vn, tf 1, . . . , <7n, t), Qa = Qa(Vu • • • , Pn, 01, . . . , q n , t) (14) 

denoted briefly by 

Pa = Pa (Pa, <?a, t), Qa = Qa(Pa, Qa, t) (15) 

We restrict ourselves to transformations called canonical or contact transformations for 
which there exists a function Jl called the Hamiltonian in the new coordinates such that 

P «--Wa> Qa ~^Pa {W) 

In such case we often refer to Q a and P a as canonical coordinates. 

The Lagrangians in the old and new coordinates are L(p a ,q a ,t) and oC(P a ,Qa,t) re- 
spectively. They are related to the Hamiltonians H(p a ,q a ,t) and Jl(P a ,Q a ,t) by the 
equations 

H = 2 Paka ~ L, Ji = 2 PaQa ~ * (17) 

where the summations extend from a = 1 to n. 



CONDITION THAT A TRANSFORMATION BE CANONICAL 

The following theorem is of interest. 
Theorem 12.2. The transformation 

Pa = Pa(Pa,qa,t), Qa = Qa(Pa,q a ,t) (18) 

is canonical if ^p a dq a — ^PadQa (I 9 ) 

is an exact differential. 

GENERATING FUNCTIONS 

By Hamilton's principle the canonical transformation (14) or (15) must satisfy the con- 
ditions that f * L dt and f ' -C dt are both extrema, i.e. we must simultaneously have 

8 C* Ldt = and sf'^f = (20) 



These will be satisfied if there is a function Q such that 

4? = L - JZ («) 

dt 

See Problem 12.11. We call Q a generating function. 

By assuming that Q is a function, which we shall denote by d", of the old position co- 
ordinates q a and the new momentum coordinates P a as well as the time t, i.e., 



CHAP. 12] HAMILTONIAN THEORY 315 

Q = effaa Pa, t) (22) 
we can prove that [see Problem 12.13] 

where P. = -M q. = M (w) 

Similar results hold if the generating function is a function of other coordinates [see 
Problem 12.12]. 



THE HAMILTON-JACOBI EQUATION 

If we can find a canonical transformation leading to Jl = 0, then we see from (24) that 
P a and Q a will be constants [i.e., P a and Q a will be ignorable coordinates]. Thus by means of 
the transformation we are able to find p a and q a and thereby determine the motion of the 
system. The procedure hinges on finding the right generating function. From the third 
equation of (23) we see by putting Jl = that this generating function must satisfy the 
partial differential equation 

|^- + H(p a , q a , t) = (25) 

f + *(f. «-.«) = ° <«•> 

This is called the Hamilton-Jacobi equation. 



SOLUTION OF THE HAMILTON-JACOBI EQUATION 

To accomplish our aims we need to find a suitable solution of the Hamilton-Jacobi 
equation. Now since this equation contains a total of n + 1 independent variables, i.e. 
Qu qz, . . . , q n and t, one such solution called the complete solution, will involve n + 1 con- 
stants. Omitting an arbitrary additive constant and denoting the remaining n constants by 
Pu (32, . . . , p n [none of which is additive] this solution can be written 

of = offai, tf2, ... , q n , pu fa, ..., £„, *) (27) 

When this solution is obtained we can then determine the old momentum coordinates by 

def 

Also, if we identify the new momentum coordinates P a with the constants p a , then 

M 

dfia 



O — doJ* _ 
where y a , a = 1, . . . , n are constants. 

Using these we can then find q a as functions of p a , y a and t, which gives the motion of 
the system. 



CASE WHERE HAMILTONIAN IS INDEPENDENT OF TIME 

In obtaining the complete solution of the Hamilton-Jacobi equation, it is often useful to 
assume a solution of the form 



316 HAMILTONIAN THEORY [CHAP. 12 

of = Si(qi) + -S 2 (g 2 ) + • • • + Sn{q n ) + F(t) (30) 

where each function on the right depends on only one variable [see Problems 12.15 and 
12.16]. This method, often called the method of separation of variables, is especially useful 
when the Hamiltonian does not depend explicitly on time. We then find that F(t) = —Et, 
and if the time independent part of of is denoted by 

5 = Si(qi) + S2M + ••■ + S n {q n ) (31) 

the Hamilton-Jacobi equation (26) reduces to 

*(£•«■) = E (S2) 

where E is a constant representing the total energy of the system. 

The equation (32) can also be obtained directly by assuming a generating function S 
which is independent of time. In such case equations (23) and (24) are replaced by 

Pa = |S , Q a = j§-, JH = H = E (33) 

dq a OJTa 

where Pa = ~tq~> Q<* — Tp~ (**) 



PHASE INTEGRALS. ACTION AND ANGLE VARIABLES 

Hamiltonian methods are useful in the investigation of mechanical systems which are 
periodic. In such case the projections of the motion of the representative point in phase 
space on any p a q a plane will be closed curves C«. The line integral 

J a = <f> Padq a ( 35 ) 

J c a 

is called a phase integral or action variable. 

We can show [see Problems 12.17 and 12.18] that 

S = S(qi, ...,qn,Ju ...,Jn) (*6) 

where p. = Jg, «« = |J ^ 

It is customary to denote the new coordinates Q a by w a so that equations (37) are re- 

pIacedby as bs (m 

Thus Hamilton's equations become [see equations (33) and (34)] 

dW a dJ a 

where Jl = E in this case depends only on the constants J a . Then from the second equa- 
tion in (39), nn s 

V } Wa = fat + Ca (*°) 

where /« and c a are constants. We call w a angle variables. The frequencies /« are given by 

dJa 

See Problems 12.19 and 12.20. 



CHAP. 12] 



HAMILTONIAN THEORY 317 



Solved Problems 



THE HAMILTONIAN AND HAMILTON'S EQUATIONS 

12.1. If the Hamiltonian H = 2 V«qa - L, where the summation extends from 
a = 1 to n, is expressed as a function of the coordinates q a and momenta p a , prove 
Hamilton's equations, ^ vrr ^ d jj 

V" = 'Wee' *" = W* 
regardless of whether H (a) does not or (b) does contain the variable time t explicitly. 

(a) H does not contain t explicitly. 

Taking the differential of H - 2 Va Qa — L > we have 

dH = 2P a <*<7« + 2?«# a - ^^dq a - 1jr-di a (1) 

Then using the fact that p a = dL/dq a and p a = dL/dq a , this reduces to 

dH - *2,q a dp a - 2p a dq a {2) 

But since H is expressed as a function of p a and q a , we have 

dH = 2jr-dp a + 1^-dq a (3) 

dp a dq a 

Comparing (2) and (3) we have, as required, 

. _ dH_ . BH 

q <* - dp a > Pa dq a 

(b) H does contain t explicitly. 

In this case equations (1), (2) and (3) of part (a) are replaced by the equations 

dH = 2?>a<4* + 2g«<*P« - 2f^<*<7« - 2^-<*<L - ~^dt U) 

dH = 2q a dp a - *2>p a dq a - — dt (5) 

m = 2 £ dp „ + 2 f|-^ + ff ,* « 

Then comparing (5) and (6), we have 

. _ dH_ . _ _d# 8H _ _3L 

9 « ~ dp a ' Pa ~ dq tt ' dt dt 

12.2. If the Hamiltonian H is independent of t explicitly, prove that it is (a) a constant and 
is (b) equal to the total energy of the system. 

(a) From equation (2) of Problem 12.1 we have 

-jfe = 2 ka Pa ~ 2 Pa Qa = 

Thus H is a constant, say E. 
(6) By Euler's theorem on homogeneous functions [see Problem 11.47, page 305], 



dq a 



318 



HAMILTONIAN THEORY 



[CHAP. 12 



where T is the kinetic energy. Then since p a - dL/dq a = dT/dq a [assuming the potential V 
does not depend on q a ], we have 2p a <Za = 2T. Thus as required, 

H = 2,PaQa~L = 2T-(T-V) = T + V = E 



12.3. A particle moves in the xy plane under the influence of a central force depending only 
on its distance from the origin, (a) Set up the Hamiltonian for the system. (6) Write 
Hamilton's equations of motion. 

(a) Assume that the particle is located by its polar coordinates (r, e) and that the potential due to 
the central force is V(r). Since the kinetic energy of the particle is T = -£m(r 2 + r 2 * 2 ), the 

Lagrangian is .„„•„,,-,, „ 

L = T - V = im(r 2 + rW) - V(r) 



We have 
so that 



Pr 



dL/dr = mr, p g = dL/de — mr 2 6 
r = p r /m, o = p e /wr 2 



Then the Hamiltonian is given by 



H = £ p a q« - L = P r r + PeO - {$m(r* + r&) - V(r)> 



= Pr S +" 3 - 4* 



Pe 



P!. + r2 .^LV V(r) 
m 2 m 2 W * ' 



2m 2mr 2 ' 

Note that this is the total energy expressed in terms of coordinates and momenta. 

(6) Hamilton's equations are q a = dH/dp a , p a = -dH/dq a 

Thus ; = dH/dp r = p r lm, h = dH/dpe - p Q /mr* 

p r = -dH/dr = ppmr* - V(r), p e = -dH/de = 
Note that the equations (5) are equivalent to the corresponding equations (3). 



(1) 
(2) 
(3) 



(4) 



(5) 
(6) 



PHASE SPACE AND LIOUVILLE'S THEOREM 

12.4. Prove Liouville's theorem for the case of one degree of freedom. 



We can think of the mechanical system as 
being described in terms of the motion of rep- 
resentative points through an element of vol- 
ume in phase space. In the case of a mechanical 
system with one degree of freedom, we have a 
two dimensional (p, q) phase space and the vol- 
ume element reduces to an area element dpdq 
[Fig. 12-2]. 

Let p = p(p, q, t) be the density of rep- 
resentative points, i.e. the number of repre- 
sentative points per unit area as obtained by 
an appropriate limiting procedure. Since the 
speed with which representative points enter 
through AB is q, the number of representative 
points which enter through AB per unit time is 

pq dp (1) 

The number of representative points which leave through CD is 



B(q, p + dp) C(q + dq, P + dp) 



#■ * 

.. I »» 



Mq,v) 



Fig. 12-2 



P q + j- (pq) dq f dp 



D(q + dq, p) 



CHAP. 12] HAMILTONIAN THEORY 319 

Thus the number which remain in the element is (i) minus (2), or 

—^(pq)dpdq (3) 

Similarly the number of representative points which enter through AD and leave through BC are 
respectively , -. 

pp dq and <pp + — (pp) dp > dq 

Thus the number which remain in the element is 

- — (pp)dpdq (-4) 

The increase in representative points is thus [adding (3) and (4)] 



{*#+^V** 



Since this is equal to -^dpdq, we must have 



dp + [ Kpq) + d(pfr ) 
dt 1 dq dp 

dp , dq , dp • , dp . op . . 

tt + PF"+ -f-q + P-^- + -r-p = (5) 

dt dq dq * dp dp v 



Now by Hamilton's equations p = —dH/dq, q = dH/dp so that 

dp _ dm dq _ dm 



dp dpdq' dq dq dp 

Thus since we suppose that the Hamiltonian has continuous second order derivatives, it follows 
that dp/dp = —dq/dq. Using this in (5), it becomes 

But this can be written dp/dt = (7) 

which shows that the density in phase space is constant and thus proves Liouville's theorem. 



12.5. Prove Liouville's theorem in the general case. 

In the general case the element of volume in phase space is 

dV = dq y • • • dq n dp t • • • dp n 

In exactly the same manner as in Problem 12.4 the increase of representative points in dV is found 
to be 

(3(p«i) , , d(pq n ) a(pp t ) d(pP n )\ _. 

— < — (-•••+ — 1- — r 1- • • • H > dV 

{ dq t dq n 3p! dp n J 

and since this is equal to -£dV, we must have 
dt 

dp , d( pqi ) d(pq n ) d( pPl ) d(pp n ) 

—- -f- — _ \. . . . _|_ — -J- _^ -j. . . . _|_ — _ y 

dt dq x dq n dp t dp n 

dp , £ a <Pg«) . x d ^) 
dt a =l dq a «=i dp a 

This can be written as 

dp , £ (dp . , dp • \ , « fdq a ,dp a \ 



320 HAMILTONIAN THEORY [CHAP. 12 

Now by Hamilton's equations p a = —dH/dq a , q a = dH/dp a so that 

dp a _ d 2 H dq a _ d 2 H 

^K ~~ ~ d P<x d<la ' dPa ~ d <I a d Pa 

Hence dp a /dp a = -dq a /dq a and (1) becomes 

&+ 2 (£-«« + s^O = o («) 

at a=l \^a "Pa / 

i.e., <V<tt = (#) 

or p = constant. 

Note that we have used the fact that if p = p(q u . . . , q n > Pi, ■ ■ ■ > Pn> *) then 

dp 4j / dp dq a dp dp a \ dp _ » fdp_. ,dp_- \ , dp 



CALCULUS OF VARIATIONS AND HAMILTON'S PRINCIPLE 

F(a, y, y') dx to be an extremum [maxi 
mum or minimum] is 



d L (dF_\_dF = Q 



dx \ by' ) dy 
Suppose that the curve which makes I an extremum is given by 

y = Y(x), a^x^b CO 

Then V = Y(x) + e V (x) = Y + e V <*) 

where e is independent of x, is a neighboring curve through x = a and x = b if we choose 

via) = v(b) = W 

The value of 7 for this neighboring curve is 

F(x, Y + ev , Y' + e V ') dx U) 

a 

jr I 

This is an extremum for e = 0. A necessary condition that this be so is that -^ | e=o = 0. But by 
differentiation under the integral sign, assuming this is valid, we find 

de | £ =o J a \dy v ^ dy' v J 

which can be written on integrating by parts as 

- £'{£-=(5)}* - ° 

where we have used (3). Since v is arbitrary, we must have 

dF _d_/dF\ _ n nr ±{§L\-.W = o 
~dy dx\dy'J ~ u dx\dy'J dy 

which is called tf wtor** or Lagrange's equation. The result is easily extended to the integral 

Xb 
F{X, i/i, If J, 2/2» 1/2. • • ■ » Vn» »n) d * 
u 

and leads to the Euler's or Lagrange's equations 



CHAP. 12] 



HAMILTONIAN THEORY 



321 



dx\by' a J dy a 



a = 1,2, ...,n 



By using a Taylor series expansion we find from (4) that 



/( e ) _ j(o) = e j f j- v + —T)')dx + higher order terms in e 2 , e 3 , etc. 

The coefficient of e in (5) is often called the variation of the integral and is denoted by 

F(x, y, y') dx 



(5) 



■f. 

F(x, y, y') dx is an extremum is thus indicated by 
a b 

8 f F(x, y, y') dx = 

12.7. Discuss the relationship of Hamilton's principle with Problem 12.6. 

By identifying the function F{x,y,y') with the Lagrangian L(t,q,q) where x,y and y' are re- 
placed by t, q, q respectively, we see that a necessary condition for the action integral 



r 



Ldt 



(1) 



to be an extremum [maximum or minimum] is given by 

l(f)-f = ° < 2) 

Since we have already seen that (2) describes the motion of a particle, it follows that such motion 
can also be achieved by requiring that (1) be an extremum, which is Hamilton's principle. 

For systems involving n degrees of freedom we consider the integral (1) where 

L = L(t, q v q lt q 2 , q 2 , . . . , q n , q n ) 

which lead to the Lagrange equations 



dt\dq a J dq a 



= 



a = 1,2, . . .,« 



A(« »Vo) 



12.8. A particle slides from rest at one point on a 
f rictionless wire in a vertical plane to another 
point under the influence of gravity. Find the 
total time taken. 

Let the shape of the wire be indicated by curve 
C in Fig. 12-3 and suppose that the starting and fin- 
ishing points are taken to be the origin and the point 
A{xQ,yo) respectively. 

Let P(x, y) denote any position of the particle 
which we assume has mass m. From the principle 
of conservation of energy, if we choose the horizontal 
line through A as reference level, we have 

Potential energy at O + kinetic energy at O = potential energy at P + kinetic energy at P 
or mgy + = mg(y — y) + %m(ds/dt) 2 

where ds/dt is the instantaneous speed of the particle at time t. Then 

ds/dt = ±Vzgy (-0 

If we measure the arc length s from the origin, then s increases as the particle moves. Thus ds/dt 
is positive, so that ds/dt = y/2gy or dt = ds/y/2gy. 

The total time taken to go from y = to y = y is 




Fig. 12-3 



322 HAMILTONIAN THEORY [CHAP. 12 

T = I dt = I 

^0 ^=0 



'2gy 



But (cte) 2 = (do;) 2 + (dy) 2 or cte = ^1 + y> 2 dx. Thus the required time is 

1 r v Vi + y' 2 

r = —— I j=- dx (2) 

V2gJ y=0 vv 

12.9. If the particle of Problem 12.8 is to travel from point O to point A in the least pos- 
sible time, show that the differential equation of the curve C denning the shape of 
the wire is 1 + y' 2 + lyy" = 0. 

A necessary condition for the time r given by equation (2) of Problem 12.8 to be a minimum 
is that 

±(d_F_\_8F = 

dx\dy'J dy U K1) 

where F = (1 + y' 2 ) 1 ' 2 V V2 (2) 

Now BFlby' = (l + y' 2 )- 1/2 y'y-^ 2 , dF/dy = — 1(1 + y' 2 )^ y -s/2 

Substituting these in (1), performing the indicated differentiation with respect to x and simplifying, 
we obtain the required differential equation. 

The problem of finding the shape of the wire is often called the brachistochrone problem. 

12.10. (a) Solve the differential equation in Problem 12.9 and thus (b) show that the required 
curve is a cycloid. 

(a) Since x is missing in the differential equation, let y' = u so that 

„ _ du _ dw dy_ _ du , _ du 
dx dy dx dy dy 

Then the differential equation becomes 

l + u* + 2yu ( ^=0 or -^ + ^ = 
dy 1 + u z y 

Integration yields 

In (1 + u 2 ) + In y = In 6 or (1 + u 2 )y = b 

where 6 is a constant. Thus 



- = * = t - J ^ 



\ y 

since the slope must be positive. Separating the variables and integrating, we find 

Letting y — b sin 2 e, this can be written 

x = f J b sin2 9 . 26 sin ecosede + c 
J V 6 cos 2 e 

= 26 f sin 2 6 de + c = 6 j (1 - cos 2e) de + c = £6(2* - sin 2e) + c 

Thus the parametric equations of the required curve are 

x = £6(2* - sin 2d) + c, y = b sin 2 e = |6(1 - cos 2e) 

Since the curve must pass through the point x = 0, y = 0, we have c = 0. Then letting 

= 2(9, a = |6 CO 

the required parametric equations are 

x = a(<p — sin <f>), y = a(l — cos <p) (2) 



CHAP. 121 



HAMILTONIAN THEORY 



323 



(b) The equations (2) are parametric equations of a cycloid [see Fig. 12-4]. The constant a must 
be determined so that the curve passes through point A. The cycloid is the path taken by a 
fixed point on a circle as it rolls along a given line [see Problem 12.89]. 



7< 
/ \ 




CANONICAL TRANSFORMATIONS AND GENERATING FUNCTIONS 
12.11. Prove that a transformation is canonical if there exists a function q such that 
dq/dt = L-Ji. 

The integrals I Ldt and I £dt must simultaneously be extrema so that their varia- 



tions are zero, i.e., 



Thus by subtraction, 



t. 



! \ Ldt - and SJ -C dt = 



f 



(L - £) dt 







This can be accomplished if there exists a function Q such that 

L - -C = d£/ctt 

-r-dt = S{£(£ 2 ) — £(*i)} = o 
The function Q is called a generating function. 



12.12. Suppose that the generating function is a function T of the old and new position co- 
ordinates q a and Q a respectively as well as the time t, i.e. T = T(q a , Q a , t). Prove that 



Va = ||-, P a = -*[1, ^ = ^ + tf where P« = 



dQ a ' 



dt 



dQa ' dPa 



By Problem 12.11, 



dT 
dt 



= L-j£ = 2, Pa q a - H - V2P a Q a - Si 

= Sp«g« ~ 2P«Q a + Si - H 

dT = 2p a dq a - 2P a dQ a + (Si~H)dt 



But if T = T{q a , Q a , t), then 



39o 

Comparing (1) and (2), we have as required 

dT 



dT = 2£*Ia+ 2^dQ a + £dt 



3Q« 



dt 



Pa - 



dq a ' 



P„ = - 



3t 



*c# 



- _ djj 
Qa ~ 8P„ 



(1) 



The equations * « — aQ ' 

follow from the fact that Si is the Hamiltonian in the coordinates P a , Q a so that Hamilton's equa- 
tions hold as in Problem 12.1. 



324 HAMILTONIAN THEORY [CHAP. 12 

12.13. Let efbea generating function dependent only on q a , P a , t. Prove that 

Va ~W«' Qa = eK' J( = lt +H where Pa = -Wa' Qa = W~ a 

From Problem 12.12, equation (1), we have 

dT = ^p a dq a - ^,P a dQ a + (J(-H)dt 



Or ^T + ^PaQa) = 2Pa^a + 2 Q« <*P« + (Jl~H)dt (1) 

i.e., def = 2p«dg a + 2Q«dP a + (J(-H)dt (2) 

where gf = T + ^P a Q a (3) 

But since gf is a function of q a , P a , t, 

^ = 2^^«+ SJ^p. + J^ «) 

Comparing (2) and (4), 3 , fl f a , 

The results P a = -777-, Q a = 75- 

follow as in Problem 12.12, since J# is the Hamiltonian. 



12.14. Prove that the transformation P = i(p 2 + q 2 ), Q = tan -1 (q/p) is canonical. 
Method 1. 

Let the Hamiltonians in the coordinates p, q and P, Q be respectively H(p, q) and J#(P, Q) so 
that H(p, q) — Jfl(P, Q). Since p, q are canonical coordinates, 

• dH • 3jF/ /^x 

P = —~r- , Q - 7— (■*) 

dq dp 

; = i^H4, i = *> + *« » 

dH ^ 9J(dP dj£dQ dH = ?MdP 3JtdQ () 

dq dP dq dQ dq' dp dP dp dQ dp 

From the given transformation equations we have 

5P _ dP __ dQ __ -q dQ _ p 

dp ~ V ' dq ~ q ' dp p 2 + q 2 ' dq p 2 + q 2 

Also, differentiating the transformation equations with respect to P and Q respectively, we find 

* = »§ + <§• • = (>&-<&)A + «*> 

Solving simultaneously, we find 

dp __ P dq. - 9. *L - - a M. = v U) 

dP " p 2 + q 2 ' dP ~ p 2 + q 2 ' dQ q ' dQ V 



CHAP. 12] HAMILTONIAN THEORY 325 

Then equations (1) and (2) become 

sh _ djj p djj en BJj q Bjj 

dq ~ Q dP P 2 + q 2 dQ ' dp P dP p 2 + q 2 dQ K ' 



Thus from equations (1), (5) and (6) we have 



P 



qQ 



sji p ajl 



p 2 + q 2* ™ H 8P p 2 + q 2 dQ 

q JL . A 9Jl Q dj( 



; P + pQ = p 



p 2 + g 2* r ** ~ * d p p 2 + q2 dQ 
Solving these simultaneously we find 

. _ _ej( . _ 3_M 

F " dQ ' * ~ dP 
which show that P and Q are canonical and that the transformation is therefore canonical. 



(7) 



Method 2. 

By Theorem 12.2, page 314, the transformation is canonical if 

1p a dq a - J,P a dQ a (8) 

is an exact differential. In this case (8) becomes 

pdq - PdQ = pdq - ±(p 2 + q 2 ) ( Pdq-qdp 

\ P + 9 

= %(P dq + q dp) = d{±pq) 
an exact differential. Thus the transformation is canonical. 



THE HAMILTON-JACOBI EQUATION 

12.15. (a) Write the Hamiltonian for the one dimensional harmonic oscillator of mass m. 
(b) Write the corresponding Hamilton-Jacobi equation, (c) Use the Hamilton-Jacobi 
method to obtain the motion of the oscillator. 

(a) Method 1. 

Let q be the position coordinate of the harmonic oscillator, so that q is its velocity. Since 
the kinetic energy is T = ^mq 2 and the potential energy is V = ^*cg 2 , the Lagrangian is 

L = T - V = \mq 2 - \ K q 2 (1) 

The momentum is p = dL/dq = mq (2) 

so that q = p/m (3) 

Then the Hamiltonian is 

H = *2,p a q a - L = pq - (%mq 2 - % K q 2 ) 

= £p 2 /m + % K q 2 U) 

Method 2. 

By Problem 12.2, since the Hamiltonian is the same as the total energy for conservative 
systems, 

H = ±mq 2 + \nq 2 — \m{plm) 2 + £«<7 2 = \p 2 lm + ±ieq 2 



326 HAMILTONIAN THEORY [CHAP. 12 

(b) Using p = dof/dq and the Hamiltonian of part (a), the Hamilton-Jacobi equation is [see equa- 
tion (26), page 315] 

(c) Assume a solution to (5) of the form 

ef = SM + S 2 (t) (6) 

1 /dS x \ 2 dS 2 

Then (5) becomes 2^W7 + ^ = ~~dt (7) 

Setting each side equal to the constant /?, we find 

whose solutions, omitting constants of integration, are 

Si = f V2m( i 8 - l/cg 2 ) ^ s 2 = -pt 

so that (6) becomes ^ = j V2m(/3 - ^/cg 2 ) dq ^ pt 

Let us identify /? with the new momentum coordinate P. Then we have for the new position 
coordinate, 



(8) 



(9) 



_ B<£ _ _§_ 
Q ~ dp dp 



\ f y/2m(p - ^/cg 2 ) dq - 0* 
V2m r dq 



C dq 

J -v/rt — 



2 J y/p - %Kq* 
But since the new coordinate Q is a constant y, 

ViJm /" dq 



- t 



s 



y/p - &Q 2 



- t = y 



or on integrating, VW* sin" 1 (qy/ic/2p ) - t + y 

Then solving for g, q = y/2plic sin yJTJm (t + y) (-* ) 

which is the required solution. The constants p and y can be found from the initial conditions. 

It is of interest to note that the quantity p is physically equal to the total energy E of the 

system [see Problem 12.92(a)]. The result (9) with p = E illustrates equation (SI) on page 316. 



12.16. Use Hamilton-Jacobi methods to solve Kepler's problem for a particle in an inverse 
square central force field. 

The Hamiltonian is H = ^ \ p ? + ~^ J ~ ~r 



Then since p r = d^/dr, p e = d^/de, the Hamilton-Jacobi equation is 

It + 2m \V Br J ^ r 2 V Be J J 
Let J = S 1 (r) + S 2 (e) + S 3 (t) W 



* = 



CHAP. 12] HAMILTONIAN THEORY 327 

1 f/dSA 2 , 1 f dS A\ K _ dS s 

Then (2) becomes ^ ^r J + ^{^fj J " ~ ~ "^ 

Setting both sides equal to the constant /? 3 , we find 

dS z ldt = -/? 3 (4) 

^{(^) 2 ^(f )} - f - * 

Integration of (4) yields, apart from a constant of integration, 

S 3 = -/? 3 * 

Multiply both sides of (5) by 2mr 2 and write it in the form 

-dv) = r2 \ 2m ^ + -r-{-dir 

Then since one side depends only on e while the other side depends only on r, it follows that each 
side is a constant. Thus 

dS 2 /de = p 2 or S 2 = P 2 (0) 



r \ dr 
dS t 



(8) 
(9) 



ui -j- = V2m/3 3 + 2mK/r - 0§/r* (7) 

on taking the positive square root. Then 

S t = f \/2m/?3 + 2raK/r - /3^/r 2 dr 

Thus gT = J V2m/? 3 + 2mtf/r - 0*/r 2 dr + /3 2 e - (3 3 t 

Identifying /? 2 and /? 3 with the new momenta P r and P e respectively, we have 

Qr = Wz = W~2 S ^ 2m ^3 + 2mK/r - /? 2 /r 2 dr + e = Yi 

Qe = §7 = af~ J* ^ 2m/?3 + 2mX/r ~ % /r2 dr ~ * = 72 

since Q r and <? fl are constants, say y 1 and y 2 . On performing the differentiations with respect to 
/3 2 and /? 3 , we find 



«/ 7"' 



Pidr 

• 2 \/2m^3 + 2mX/r - jSf/r 2 



» - 7i (JO) 



, „ = = * + 7 2 (**) 

V2m/? 3 + 2mK/r - /3 2 /r 2 

The integral in (10) can be evaluated by using the substitution r = II u, and after integrating we 
find as the equation of the orbit, 

r = 2 = («) 

1 - Vl + ZPzpymK* cos (* + B-/2 - y x ) 

The constant /? 3 can be identified with the energy E [see Problem 12.92(b)], thus illustrating equation 
{31), page 316. If E = /3 3 < 0, the orbit is an ellipse; if E = 3 = 0, it is a parabola; and if 
E — /3 3 > 0, it is a hyperbola. This agrees with the results of Chapter 5. 

The equation (11) when integrated yields the position as a function of time. 



328 HAMILTONIAN THEORY [CHAP. 12 

PHASE INTEGRALS AND ANGLE VARIABLES 

12.17. Let gT be a complete solution of the Hamilton-Jacobi equation containing the n constants 
(3i, . . . , /?„. Let J a = y p a dq a . Prove that the J a are functions of the /? a only. 

We have ^ = S 1 (q l , lt . . ., n ) + ••• + S n (q n , fi lt . . . , (3 n ) - p x t (1) 

where the constant p 1 —E, the total energy. Now 

3<2f dS a 

dq a dq a v ; 

Thus J a = § Padq a = §~dq a {S ) 

But in this integration q a is integrated out, so that the only quantities remaining are the con- 
stants Pi, . . ., p n . Thus we have the n equations 

J a = J a (fiv • ■ ■ > Pn) a = l,...,n (4) 

Using (4) we can solve for p u , ..,p n in terms of J u .. .,J n and express (1) in terms of the J a . 

12.18. (a) Suppose that the new position and momentum coordinates are taken to be w a and 
J a respectively. Prove that if Ji is the new Hamiltonian, 

J a = -dM/dWa, W a = dJl/dJ a 

(b) Deduce from (a) that 

Ja = constant and w a — fat + c a 
where f a and c a are constants and f a = dJl/dJ a . 

(a) By Hamilton's equations for the canonical coordinates Q a ,P„, 

Pa = -dJ(/dQa, Qa = dJ{/dP a (1) 

Then since the new position and momentum coordinates are taken as Q a — w a and P a = J a > 
these equations become 

J a = -dJl/dWa, W a = dJl/dJ a (2) 

(b) Since ^[ — E, the new Hamiltonian depends only on the J a and not on the w a . Thus from (2) 
we have 

J a = 0, w a = constant = f a (3) 

where f a = djK/dJ a - From (3) we find, as required, 

J a = constant, w a — f a t + c a (-4) 

The quantities J a are called action variables while the corresponding integrals 

J Padq a = J a (5) 

where the integration is performed over a complete cycle of the coordinate q a , are called 
phase integrals. The quantities w a are called angle variables. 

12.19. (a) Let Aw a denote the change in w a corresponding to a complete cycle in the particu- 

lar coordinate q r . Prove that 

1 if a = r 
Aw a = i 

if a ¥^r 



CHAP. 12] HAMILTONIAN THEORY 329 

(b) Give a physical interpretation to the result in (a). 

9 C dS , dJ r fl if a = r 

'- M- a yw r dqr = — = 



(a) Aw a 



5 J a [ if a # r 

where we have used the fact that w a = aS/d«7 a [see Problems 12.17 and 12.18] and have as- 
sumed that the order of differentiation and integration is immaterial. 

(6) From (a) it follows that w a changes by one when q a goes through a complete cycle but that 
there is no change when any other q goes through a complete cycle. It follows that q a is a 
periodic function of w a of period one. Physically this means that the f a in equation (4) of 
Problem 12.18 are frequencies. 



12.20. Determine the frequency of the harmonic oscillator of Problem 12.15. 

A comple te cy cle of the c oordin ate q [see equation (10), Problem 12.15] consists in the motion 
from q = -y/2(3/ K to q = +V2/3//C and back to q = -y/2ph. Then the action variable is 



= J pdq = 2 J y/2m(/3 - \ K q*) dq = 4 J y/2m(p - % K q*) dq 

-V2/3/K 

' = * = £>£ = .* «* ' = °4 = i{i 



= 27T/?VW/C 

Thus 



12.21. Determine the frequency of the Kepler problem [see Problem 12.16]. 

^ A complete cycle of the coordinate r consists in the motion from r = r min to r max and back to 
r ~ r min> where r min and r max are the minimum and maximum values of r given by the zeros 
of the quadratic equation [see equation (10), Problem 12.16] 

2m/? 3 + 2mK/r - ppr 2 = (1) 

We then have from equations (6) and (7) of Problem 12.16, 

J 6 = § Ve de = § d ^-ds = §^de = f* fi%d , = u Pt (f) 

r r d<of rds< r rmax . 

J r = J> Pr dr = y-^rdr = J-j-dr = 2 J V2m/? 3 + 2mK/r - &/r* dr 

r min 
= 2wmK/y/-2m/3 3 - 2tt/3 2 ^ 

From (2) and (3) we have on elimination of p 2 , 

J e + J r = 2TrimKly]— 2m/3 3 m 

Since p 3 = E, (4) yields 

_ 2^2 m g2 2^2 m /f2 

^-"OVfT^ -othat jr = -__ 

Then the frequencies are 

J6 dJ 9 (./«, + J r )3' fr dJ r ~ (J e +«7 r )3 

Since these two frequencies are the same, i.e. there is only one frequency, we say that the 
system is degenerate. 



330 HAMILTONIAN THEORY [CHAP. 12 

MISCELLANEOUS PROBLEMS 

12.22. A particle of mass m moves in a force field of potential V. Write (a) the Hamiltonian 
and (b) Hamilton's equations in spherical coordinates (r, 9, </>). 

(a) The kinetic energy in spherical coordinates is 

T = \m(r 2 + r 2 e 2 + r 2 sin 2 9 2 ) (1) 

Then the Lagrangian is 

L = T -V = \m(r 2 + r 2 e 2 + r 2 s i n 2 tf £2) _ y( r , ^ ^ ^ 

We have 

p r — dL/dr = mf , p e = dL/do = mr 2 8, p^, = 3L/30 — mr 2 sin 2 (5) 

, Pr • Pe • P<*> 

and r = — , = — „-, = — 2 . 2 „ (4) 

m mr mr sin 2 v ' 

The Hamiltonian is given by 

H = ^tP a q a — L 

= p r r + p e e + pqQ — ^m(r 2 + r 2 e 2 + r 2 sin 2 e 2 ) + V(r, e, 0) 

n 2 m 2 « 2 

_ Pr_ , Pe , P<& v , , 

2m 2mr 2 2mr 2 sin 2 <? + v &>'>*) i 5 ) 

where we have used the results of equations (4). 

We can also obtain (5) directly by using the fact that for conservative systems the 
Hamiltonian is the total energy, i.e. H = T + V. 

(b) Hamilton's equations are q a — - — , p a = — t — . Then from part (a), 

dp a dq a 

dH Pe . dH P4 



. _ dH_ _ Pr • _ dH_ _ _Po_ . _ dH_ _ 



dp r m ' dp e mr 2 ' dp$ mr 2 sin 2 

* = _^ = -EL + p * _ §Z 

r dr mr 3 mr 3 sin 2 e dr 

. _ _dH_ _ P% cos dV 
P° ~ do ~ mr 2 sin 3 e do 

' = -^R - _§¥_ 

P * ~~ 30 30 

12.23. A particle of mass m moves in a force field whose potential in spherical coordinates 
is V — — (K cos 0)/r 2 . Write the Hamilton- Jacobi equation describing its motion. 

By Problem 12.22 the Hamiltonian is 

N - 1 ( a * I P ° I P * ) Kc0Sff (1) 

n ~ 2m V r r 2 r 2 sin 2 0/ r 2 K ' 

AqJ" 3<of 9(of 

Writing p r — -r— , p e = -r— , p^ — — — , the required Hamilton- Jacobi equation is 

M-l. 1 iYMY. ^^Vi 1 /^Y) gcosg = 

3* + 2m U 3r- / r 2 V 3* / r- 2 sin 2 \ 30 J r 



12.24. (a) Find a complete solution of the Hamilton- Jacobi equation of Problem 12.23 and 
(b) indicate how the motion of the particle can be determined. 

(a) Letting ©f = S x (r) + S 2 (&) + £3(0) ~ Et in equation (2) of Problem 12.23, it can be written 



CHAP. 12] HAMILTONIAN THEORY 331 



(1) 



i fds t y , i /ds 2 y i /ds 3 y Kcos9 

2m\dr J r 2mr 2 \ do J 2rar 2 sin 2 e\d<f>) r 2 ~ 

Multiplying equation (1) by 2mr 2 and rearranging terms, 

/rfSiV /dS 2 \ 2 1 /dSo\2 

Since the left side depends only on r while the right side depends on e and <f>, it follows that 
each side must be a constant which we shall call /?j. Thus 

\d7) ~ 2mEr2 = Pl (2) 

/dS 2 \ 2 1 /dSo\2 

and -(^j -a^^j +fcOr««. = fc W 

Multiplying equation (5) by sin 2 * and rearranging terms, 

/dS 3 y /dS a \* 

\~d&) = 2mKain20cos9 - /^sin 2 * — sin 2 * ( —j— J (4) 

Since the left side depends only on <p while the right side depends only on e each side must be a 
constant which we can call /3 2 . However, since 

^ = 1? = ~dj < 5 > 

we can write /3 2 = p^. This is a consequence of the fact that is a cyclic or ignorable co- 
ordinate. Then (4) becomes 

2mK sin 2 e cos 6 — p x sin 2 e - sin 2 e ( ~ ) = p| (6) 

By solving equations (2), (6) and (5), we obtain 

s i - J y/2mE + pjr 2 dr, S 2 = J yJImK cos - p 2 esc 2 * - /?i ete, S 3 = p^ 

where we have chosen the positive square roots and omitted arbitrary additive constants. The 
complete solution is 

of = J V2mE~+JJr^dr + J v / 2m J K' cos * - p$ esc 2 e - /3 1 de + p^ - #< 

(6) The required equations of motion are found by writing 

30J ~ Yl ' ~dE ~ l2 ' dp^ = Y3 

and then solving these to obtain the coordinates r, 0,0 as functions of time using initial condi- 
tions to evaluate the arbitrary constants. 

12.25. If the functions F and G depend on the position coordinates q a , momenta p a and time 
t, the Poisson bracket of F and G is denned as 

\f G] = y (**L!*L - dF dG ) 

o \dp*dq a dq a dpaj 

Prove that (a) [F,G] = -[G,F], (6) [Fi + F 2 ,G] = [F U G] + [F*,G], (c) [F,q r ] = 
dF/dpr, (d) [F,p r ] = -dF/dq r . 

(a) \F,G\ = ^(**L*G-*LdG\ _ _^(dG_dL_dGdF_\ _ _ {rp] 



a \3p« 3g a 9g a ap^/ a \dp a 3g a dq a dp a J 

This shows that the Poisson bracket does not obey the commutative law of algebra. 



332 HAMILTONIAN THEORY [CHAP. 12 

(h) ]F+F G] _ y P>(* T i + * , 2) 8G a{F l + FJ dG \ 

(b) [F 1 + F 2 ,G] - i| dpa dga dga dp ^ 

V ( d ll*<L _ M\ 8G_\ - /^2 ^G _ Mj, _3G 

= [^i,G] + [F 2 ,G] 
This shows that the Poisson bracket obeys the distributive law of algebra. 

^ f dF dq r dF dq r \ _ qF_ 
{C) [,<lrl ~ %\d Pa dq a dq a dp a J d Pr 

since dq r /dq a = 1 for a = r and for a ¥= r, while dq r /dp a = for all a. Since r is ar- 
bitrary, the required result follows. 



(d) 



[ ,Pr] ~ a\dp a dq a dq a dp a J dq r 



since dp r /dq a = for all a, while dp r /dp a = 1 for « = r and for a ¥> r. Since r is ar- 
bitrary, the required result follows. 

12.26. If H is the Hamiltonian, prove that if / is any function depending on position, momenta 
and time, then -,, «* 

or dt ~ at + r \dq a q « + dp a Pa 

dH • _ dH (*\ 

But by Hamilton's equations, q a — T~r > Pa — ~TZ~ K ' 

dp a dq a 

Then (2) can be written 

§L = U. + 2 f-^- ^L--&- —) = &-+ [H, f] 



dt dt H \dq a dp a dp a dqj dt 



Supplementary Problems 

THE HAMILTONIAN AND HAMILTON'S EQUATIONS 

12.27. A particle of mass m moves in a force field of potential V. (a) Write the Hamiltonian and (6) Ham- 
ilton's equations in rectangular coordinates (x,y,z). 

Ans. (a) H = (p| + p\ + p|)/2m + V(x, y, z) 

(b) x = pjm, y = p y /m, z = p z /m, p x = -dV/dx, p y = -dV/dy, p z = -dV/dz 

12.28. Use Hamilton's equations to obtain the motion of a particle of mass m down a frictionless inclined 
plane of angle a. 

12.29. Work the problem of small oscillations of a simple pendulum by using Hamilton's equations. 

12.30. Use Hamilton's equations to obtain the motion of a projectile launched with speed v at angle a 
with the horizontal. 



CHAP. 12] 



HAMILTONIAN THEORY 



333 



12.31. Using Hamilton's equations, work the problem of the harmonic oscillator in (a) one dimension, 
(b) two dimensions, (c) three dimensions. 

12.32. Work Problem 3.27, page 78 by uging Hamilton's equation. 

PHASE SPACE AND LIOUVILLE'S THEOREM 

12.33. Explain why the path of a phase point in phase space which represents the motion of a system of 
particles can never cross itself. 

12.34. Carry out the details in the proof of Liouville's theorem for the case of two degrees of freedom. 

CALCULUS OF VARIATIONS AND HAMILTON'S PRINCIPLE 

12.35. Use the methods of the calculus of variations to find that curve connecting two fixed points in a 
plane which has the shortest length. 

12.36. Prove that if the function F in the integral J F(x, y, y') dx is independent of x, then the integral 

a 

is an extremum if F — y'F y > — c where c is a constant. 

12.37. Use the result of Problem 12.36 to solve (a) Problem 12.9, page 322, (6) Problem 12.35. 



12.38. It is desired to revolve the curve of Fig. 12-5 hav- 
ing endpoints fixed at P(x v y x ) and Q(x 2 , y 2 ) about 
the x axis so that the area / of the surface of 
revolution is a minimum. 

/** 
yy/l + y' % dx. 

(b) Obtain the differential equation of the curve. 

(c) Prove that the required curve is a catenary. 
Ans. (6) yy" = 1 + (y')2 

12.39. Two identical circular wires in contact are placed 
in a soap solution and then separated so as to 
form a soap film. Explain why the shape of the 
soap film surface is related to the result of Prob- 
lem 12.38. 



Q(x 2 , y 2 ) 




Fig. 12-5 



12.40. Use Hamilton's principle to find the motion of a simple pendulum. 

12.41. Work the problem of a projectile by using Hamilton's principle. 

12.42. Use Hamilton's principle to find the motion of a solid cylinder rolling down an inclined plane of 
angle a. 



CANONICAL TRANSFORMATIONS AND GENERATING FUNCTIONS 

12.43. Prove that the transformation Q = p, P = —q is canonical. 

12.44. Prove that the transformation Q = q tan p, P = In sin p is canonical. 

12.45. (a) Prove that the Hamiltonian for a harmonic oscillator can be written in the form H = 4p 2 /ra + 

(b) Prove that the transformation q = \P/y/^sinQ, p - -\j mPyf^ cos Q is canonical. 

(c) Express the Hamiltonian of part (a) in terms of P and Q and show that Q is cyclic. 

(d) Obtain the solution of the harmonic oscillator by using the above results. 



334 HAMILTONIAN THEORY [CHAP. 12 

12.46. Prove that the generating function giving rise to the canonical transformation in Problem 12.45(6) 
is S — ^yficq 2 cotQ. 

12.47. Prove that the result of two or more successive canonical transformations is also canonical. 

12.48. Let V be a generating function dependent only on Q a ,p a ,t. Prove that 

12.49. Let °L> be a generating function dependent only on the old and new momenta p a and P a respectively 
and the time t. Prove that ^ w _ w 

12.50. Prove that the generating function K of Problem 12.48 is related to the generating function T of 
Problem 12.12 by V = T - 2p«<7a- 

12.51. Prove that the generating function "V of Problem 12.49 is related to the generating function T 
Problem 12.12 by "V = T + ^P a Q a - ^p a q a - 

THE HAMILTON-JACOBI EQUATION 

12.52. Use the Hamilton-Jacobi method to determine the motion of a particle falling vertically in a uniform 
gravitational field. 

12.53. (a) Set up the Hamilton-Jacobi equation for the motion of a particle sliding down a frictionless 
inclined plane of angle a. (b) Solve the Hamilton-Jacobi equation in (a) and thus determine the 
motion of the particle. 

12.54. Work the problem of a projectile launched with speed v Q at angle a with the horizontal by using 
Hamilton-Jacobi methods. 

12.55. Use Hamilton-Jacobi methods to describe the motion and find the frequencies of a harmonic oscil- 
lator in (a) 2 dimensions, (b) 3 dimensions. 

12.56. Use Hamilton-Jacobi methods to arrive at the generating function of Problem 12.46. 

PHASE INTEGRALS AND ANGLE VARIABLES 

12.57. Use the method of phase integrals and angle variables to find the frequency of a simple pendulum 

of length I, assuming that oscillations are small. An$. ^^j 

12.58. Find the frequencies of (a) a 2 dimensional harmonic oscillator, (b) a 3 dimensional harmonic 
oscillator. 

12.59. Obtain the frequency of small oscillations of a compound pendulum by using phase integrals. 



12.60. Two equal masses m connected by equal 
springs to fixed walls at A and B are free 



to slide in a line on a frictionless plane AB V A 



to slide in a line on a irictioniess piane ad y, m m ft 

[see Fig. 12-6]. Using phase integrals deter- | —nS^^-C\-^W^-C\- 'W L p| 
mine the freauencies of the normal modes. ^A ^ \^# vvv \l# £^ 



mine the frequencies of the normal modes. ^ 



1 



12.61. Discuss Problem 12.57 if oscillations are not 

assumed small. t<Ig- 



1 



MISCELLANEOUS PROBLEMS 

12 62 A particle of mass m moves in a force field having potential V( P ,*,z) where P ,*,z are cylindrical 
coordinates. Give (a) the Hamiltonian and (b) Hamilton's equations for the particle. 

Ans. (a) H = (p 2 p + p%h 2 + P 2 z )/2m + V( P , <p, z) 

(b) P = Pp/m, I = Po/mr*. z = Vz /m, p p = p%lm P * - dV/d P , P<t> = -dV/d<f>, p z = ~dV/dz 



CHAP. 121 



HAMILTONIAN THEORY 



335 



12.63. 



12.64. 



A particle of mass m which moves in a plane relative to a fixed set of axes has a Hamiltonian 
given by the total energy. Find the Hamiltonian relative to a set of axes which rotates at constant 
angular velocity «* relative to the fixed axes. 

Set up the Hamiltonian for a double pendulum. Use Hamilton-Jacobi methods to determine the 
normal frequencies for the case of small vibrations. 



12.65. 



X*2 
F{t,x,x'x)dt to be an extremum is that 



Bx dt \dx J + dV\dx) ~ ° 



12.66. 
12.67. 



BF d /BF 
Can you generalize this result? 

Work Problem 3.22, page 76, by Hamiltonian methods 



A particle of mass m moves on the inside of a f rictionless vertical cone having equation x 2 + y^ = 
* 2 tan 2 a. (a) Write the Hamiltonian and (b) Hamilton's equations using cylindrical coordinates. 

Arts. (a) H = -^- — + ir ± s + mgp cot a 



2m 

(6) } = P " Sin2a 
m 



2m P 2 



P<ft 

rap 3 



— mg cot a 



12.68. 



Use the results of Problem 12.67 to prove that there will be a stable orbit in any horizontal plane 
z — h > 0, and find the frequency in this orbit. 



12.69. 



12.70. 



Prove that the product of a position coordinate and its canonically conjugate momentum must 
have the dimension of action or energy multiplied by time, i.e. AfL 2 T-i. 

Perform the integration of equation (10) of Problem 12.16 and compare with the solution of the 
Kepler problem in Chapter 5. 



12.71. 
12.72. 

12.73. 



Verify the integration result (3) of Problem 12.21. 

Prove that Euler's equation (9), page 313, can be written as 



.n^E., „, &F 



+ 



dW 



dF 



12.74. 



12.75. 



V ' dy'2 + y dy'dy ' dy' dx by 

A man can travel by boat with speed v 1 and can walk 

with speed v 2 . Referring to Fig. 12-7, prove that in 

order to travel from point A on one side of a river 

bank to a point B on the other side in the least time 

he must land his boat at point P where angles 9 t and 

2 are such that 

sm 0j _ Vj 

sin e 2 v 2 

Discuss the relationship of this result to the refrac- 
tion of light in the theory of optics. 

Prove that if a particle moves under no external 
forces, i.e. it is a free particle, then the principle of 
least action becomes one of least time. Discuss the 
relationship of this result to Problem 12.73. Fig. 12-7 

Derive the condition for reflection of light in optical theory by using the principle of least time. 




336 



HAMILTONIAN THEORY 



[CHAP. 12 



12.76. 



It is desired to find the shape of a curve lying in a plane and having fixed endpoints such that its 

moment of inertia about an axis perpendicular to the plane and passing through a fixed origin is a 

minimum. 

(a) Using polar coordinates (r, e), show that the problem is equivalent to minimizing the integral 



- J 



r 2 V 1 + r 2 (de/dr) 2 dr 



12.77. 
12.78. 
12.79. 



where the fixed endpoints of the wire are (r v e^, (r 2 , 2 ). 

(b) Write Euler's equation, thus obtaining the differential equation of the curve. 

(c) Solve the differential equation obtained in (b) and thus find the equation of the curve. 

Ans. (c) r 3 = c t sec (30 — c 2 ) where c t and c 2 are determined so that the curve passes through 
the fixed points. 

Use the Hamilton-Jacobi method to set up the equations of motion of a spherical pendulum. 

Use Hamilton-Jacobi methods to solve Problems 11.20, page 293, and 11.21, page 294. 

If [F,G] is the Poisson bracket [see Problems 12.25 and 12.26], prove that 

(a) [F.F^G] = F t [F 2 ,G] + F 2 [F lt G\ 



<»> f t [F,G) = 



(c) 



A 

dt 



[F,G] = 



dF 

_dt 

dF_ 
dt 



,G 
,G 



+ 



+ 



w dG 



F, 



dG 
dt 



12.80. 

12.81. 
12.82. 

12.83. 
12.84. 

12.85. 

12.86. 
12.87. 



Prove that (a) [q a , q fi ] = 0, (6) [p a , p fi ] = 0, (c) [p a , q p ] = S a(3 

fl if a = fi 
where t afl = Q tf 



is called the Kronecker delta. 



Evaluate [H, t] where H is the Hamiltonian and t is the time. Are H and t canonically conjugate 
variables? Explain. 

Prove JacobVs identity for Poisson brackets 

{F lt [F 2 ,F 8 ]-] + \F 2 , [F a ,F^ + [F3,[F lf F 2 ]] = 

Illustrate Liouville's theorem by using the one dimensional harmonic oscillator. 

(a) Is the Lagrangian of a dynamical system unique? Explain. 

(6) Discuss the uniqueness of the generalized momenta and Hamiltonian of a system. 

(a) Set up the Hamiltonian for a string consisting of N particles [see Problem 8.29, page 215] 
(6) Use Hamilton-Jacobi methods to find the normal modes and frequencies. 

Prove that the Poisson bracket is invariant under a canonical transformation. 

Prove that Liouville's theorem is equivalent to the result dp/dt = [p,H], 



12.88. 



12.89. 



(a) Let Q a = 2 <*>anQ», p * = 2 *W?V where <W and 6 <*m are £ iven constants and 
a = 1, 2, . . . , n. Prove that the transformation is canonical if and only if b ail — A a(l /A 
where A is the determinant. 

«11 a 12 • • • «*ln 



a 21 a 22 



a 2n 



a n\ a n2 



a„ 



and A ajLt is the cof actor of the element a aix in this determinant. 
(6) Prove that the conditions in (a) are equivalent to the condition 2 P a Qa = 2 Pa 9a- 

Prove that the path taken by a fixed point on a circle as it rolls along a given line is a cycloid. 



CHAP. 121 



HAMILTONIAN THEORY 337 



12.90. (a) Express as an integral the total potential energy of a uniform chain whose ends are suspended 
from two fixed points. (6) Using the fact that for equilibrium the total potential energy is a mini- 
mum, use the calculus of variations to show that the equation of the curve in which the chain 
hangs is a catenary as in Problem 7.32, page 186. [Hint. Find the minimum of the integral subject 
to the constraint condition that the total length of the chain is a given constant.] 

12.91. Use the methods of the calculus of variations to find the closed plane curve which encloses the 
largest area. 

12.92. Prove that the constants (a) /? in Problem 12.15 and (b) fo in Problem 12.16 can be identified with 
the total energy. 

12.93. If the theory of relativity is taken into account in the motion of a particle of mass m in a force 
field of potential V, the Hamiltonian is given by 



H = vVc 2 + m 2 c 4 + V 
where c is the speed of light. Obtain the equations of motion for this particle. 

12.94. Use Hamiltonian methods to solve the problem of a particle moving in an inverse cube force field. 

12.95. Use spherical coordinates to solve Kepler's problem. 

12.96. Suppose that m of the n coordinates q lt q 2 , . . . , q n are cyclic [say the first ra, i.e. q x , q 2 , . . . , q m ]' Let 

TO 

% — 2 c a h a ~ L where c a = dL/dq a 
a=l 



Prove that for a = m + 1, . . .,n -77 ( tj- ) — t— - 

dt \ dq a J dq a 



The function % is called Routh's function or the Routhian. By using it a problem involving n 
degrees of freedom is reduced to one involving n — m degrees of freedom. 

rlT rlT 

12.97. Using the properties SL = — 8y + ^-7 8y', (8y)' = 8y' 

of the variational symbol 8 [see Problem 12.6] and assuming that the operator 8 can be brought 
under the integral sign, show how Lagrange's equations can be derived from Hamilton's principle. 

12.98. Let P = Pip, q), Q = Q(p, q). Suppose that the Hamiltonian expressed in terms of p, q and P, Q 
are given by H - H(p, q) and J[ = J((P, Q) respectively. Prove that if 

q = 8H/dp, p = -dH/dq 
then Q = dJl/dP, P = -dJ{/dQ 

provided that the Jacobian determinant [or briefly Jacobian] 

d(P, Q) _ dP/dp dP/dq 

Hp,q) ~ dQ/dp dQ/dq 

Discuss the connection of the results with Hamiltonian theory. 

12.99. (a) Set up the Hamiltonian for a solid cylinder rolling down an inclined plane of angle a. 

(b) Write Hamilton's equations and deduce the motion of the cylinder from them. 

(c) Use Hamilton-Jacobi methods to obtain the motion of the cylinder and compare with part (6). 

12.100. Work Problem 7.22, page 180, by Hamilton-Jacobi methods. 



= 1 



338 HAMILTONIAN THEORY [CHAP. 12 

12.101. Write (a) the Hamiltonian and (6) Hamilton's equations for a particle of charge e and mass m 
moving in an electromagnetic field [see Problem 11.90, page 309]. 

Ans. (a) H - ^-(p- eA) 2 + e<*> 

(6) v = — (p-eA), p = -eV* + eV(A«v) 

12.102. (a) Obtain the Hamilton-Jacobi equation for the motion of the particle in Problem 12.101. (6) Use 
the result to write equations for the motion of a charged particle in an electromagnetic field. 

12.103. (a) Write the Hamiltonian for a symmetrical top and thus obtain the equations of motion. (6) Com- 
pare the results obtained in (a) with those of Chapter 10. 

12.104. Prove Theorem 12.2, page 314. 



12.105. An atom consists of an electron of charge — e moving in a central force field F about a nucleus of 

F = 



charge Ze such that _ 2 



r3 

where r is the position vector of the electron relative to the nucleus and Z is the atomic number. 
In Bohr's quantum theory of the atom the phase integrals are integer multiples of Planck's con- 
stant h, i.e., ~ •» 

<j) p r dr = n^h. d) p e de — n 2 h 

Using these equations, prove that there will be only a discrete set of energies given by 

„ 2?r 2 mZ 2 e 4 

where n = n x + n 2 = 1, 2, 3, 4, . . . is called the orbital quantum number. 



Appendix A 



Units and Dimensions 



UNITS 

Standardized lengths, times and masses in terms of which other lengths, times and 
masses are measured are called units. For example, a distance can be measured in terms of 
a standard foot or meter. A time interval can be measured in terms of seconds; hours or 
days. A mass can be measured in terms of pounds or grams. Many different types of units 
are possible. However, there are four main types in use at the present time. 

1. The CGS or centimeter-gram-second system. 

2. The MKS or meter-kilogram-second system. 

3. The FPS or foot-pound-second system. 

4. The FSS or foot-slug-second system, also called the 

English gravitational or engineering system. 

The first two are sometimes called metric systems, while the last two are sometimes called 
English systems. There is an increasing tendency to use metric systems. 

The following indicates four consistent sets of units in these systems which can be used 
with the equation F = ma: 

CGS System: F (dynes) = m (grams) x a (cm/sec 2 ) 

MKS System: F (newtons) = m (kilograms) x a (m/sec 2 ) 

FPS System: F (poundals) = m (pounds) x a (ft/sec 2 ) 

FSS System: F (pounds weight) = m (slugs) x a (ft/sec 2 ) 

In the third through sixth columns of the table on page 340, units of various quantities in 
these systems are given. 

In the table on page 341, conversion factors among units of the various systems are 
given. 



DIMENSIONS 

The dimensions of all mechanical quantities may be expressed in terms of the funda- 
mental dimensions of length L, mass M, and time T. In the second column of the table on 
page 340, the dimensions of various physical quantities are listed. 

339 



340 



UNITS AND DIMENSIONS 



[APPENDIX A 



UNITS AND DIMENSIONS 



Physical Quantity 


Dimension 


CGS System 


MKS System 


FPS System 


FSS System 


Length 


L 


cm 


m 


ft 


ft 


Mass 


M 


gm 


kg 


lb 


slug 


Time 


T 


sec 


sec 


sec 


sec 


Velocity 


LT-i 


cm/sec 


m/sec 


ft/sec 


ft/sec 


Acceleration 


LT-* 


cm/ sec 2 


m/sec 2 


ft/sec 2 


ft/ sec 2 


Force 


MLT-* 


gm cm/sec 2 
= dyne 


kg m/sec 2 
= newton 


lb ft/sec 2 
= poundal 


slug ft/sec 2 
= lbwt 


Momentum, Impulse 


MLT-i 


gm cm/sec 
= dyne sec 


kg m/sec 
= nt sec 


lb ft/sec 
= pdl sec 


slug ft/ sec 
= lbwt sec 


Energy, Work 


ml* r-2 


gm cm 2 / sec 2 
= dyne cm 
= erg 


kg m 2 /sec 2 
= nt m 
= joule 


lb f t 2 /sec 2 
= ft pdl 


slug ft 2 /sec 2 
= ft lbwt 


Power 


ML 2 r-a 


gm cm 2 /sec 3 
= dyne cm/sec 
= erg/sec 


kg m 2 /sec 3 
= joule/sec 
= watt 


lb f t 2 /sec 3 
= ft pdl/sec 


slug f t 2 /sec 3 
= ft lbwt/sec 


Volume 


L 3 


cm 3 


m 3 


ft 3 


ft 3 


Density 


ML-3 


gm/cm 3 


kg/m 3 


lb/ft 3 


slug/ft 3 


Angle 


— 


radian (rad) 


rad 


rad 


rad 


Angular velocity 


T -i 


rad/sec 


rad/sec 


rad/sec 


rad/sec 


Angular acceleration 


ji-2 


rad/sec 2 


rad/sec 2 


rad/sec 2 


rad/sec 2 


Torque 


ML 2 T-2 


gm cm 2 /sec 2 
= dyne cm 


kg m 2 /sec 2 
= nt m 


lb ft 2 /sec 2 
= ft pdl 


slug ft 2 /sec 2 
= ft lbwt 


Angular momentum 


ML 2 T~i 


gm cm 2 /sec 


kg m 2 /sec 


lb f Wsec 


slug ft 2 /sec 


Moment of inertia 


Ml? 


gm cm 2 


kg m 2 


lb ft 2 


slug ft 2 


Pressure 


ML-i T-2 


gm/(cm sec 2 ) 
= dyne/cm 2 


kg/(m sec 2 ) 
= nt/m 2 


pdl/ft 2 


lbwt/ft 2 

J . 



APPENDIX A] 



UNITS AND DIMENSIONS 



341 



CONVERSION FACTORS 



Length 



Area 



Volume 



Mass 



Speed 



Density 



Force 



Energy 



Power 



Pressure 



1 kilometer (km) 

1 meter (m) 

1 centimeter (cm) 

1 millimeter (mm) 

1 micron (/t) 

1 millimicron (m/t) 

1 angstrom (A) 



1000 meters 
100 centimeters 
10-2 m 
10" 3 m 
10" 6 m 
10-9 m 
10- 10 m 



1 inch (in.) 

1 foot (ft) 

1 mile (mi) 

lmil 

1 centimeter 

1 meter 

1 kilometer 



2.540 cm 
30.48 cm 
1.609 km 
10- 3 in. 
0.3937 in. 
39.37 in. 
0.6214 mile 



1 square meter (m 2 ) = 10.76 ft 2 
1 square foot (ft 2 ) = 929 cm 2 



1 square mile (mi 2 ) = 640 acres 
1 acre = 43,560 ft 2 



1 liter (I) = 1000 cm 3 = 1.057 quart (qt) = 61.02 in 3 = 0.03532 ft 3 

1 cubic meter (m 3 ) = 1000 I = 35.32 ft 3 

1 cubic foot (ft 3 ) = 7.481 U.S. gal = 0.02832 m 3 = 28.32 I 

1 U.S. gallon (gal) = 231 in 3 = 3.785 I; 1 British gallon = 1.201 U.S. gallon = 277.4 in 3 

1 kilogram (kg) = 2.2046 lb = 0.06852 slug; 1 lb = 453.6 gm = 0.03108 slug 
1 slug = 32.174 lb = 14.59 kg 

1 km/hr = 0.2778 m/sec = 0.6214 mi/hr = 0.9113 ft/sec 
1 mi/hr = 1.467 ft/sec = 1.609 km/hr = 0.4470 m/sec 

1 gm/cm 3 = 10 3 kg/m 3 = 62.43 lb/ft 3 = 1.940 slug/ft 3 
1 lb/ft 3 = 0.01602 gm/cm 3 ; 1 slug/ft 3 = 0.5154 gm/cm 3 

1 newton (nt) = 10 5 dynes = 0.1020 kgwt = 0.2248 lbwt 

1 pound weight (lbwt) = 4.448 nt = 0.4536 kgwt = 32.17 poundals 

1 kilogram weight (kgwt) = 2.205 lbwt = 9.807 nt 

1 U.S. short ton = 2000 lbwt; 1 long ton = 2240 lbwt; 1 metric ton = 2205 lbwt 

1 joule = 1 nt m = 10 7 ergs = 0.7376 ft lbwt = 0.2389 cal = 9.481 X 10 " 4 Btu 

1 ft lbwt = 1.356 joules = 0.3239 cal = 1.285 X 10" 3 Btu 

1 calorie (cal) = 4.186 joules = 3.087 ft lbwt = 3.968 X 10~ 3 Btu 

1 Btu = 778 ft lbwt = 1055 joules = 0.293 watt hr 

1 kilowatt hour (kw hr) = 3.60 X 10 6 joules = 860.0 kcal = 3413 Btu 

1 electron volt (ev) = 1.602 X 10~ 19 joule 

1 watt = 1 joule/sec = 10 7 ergs/sec = 0.2389 cal/sec 

1 horsepower (hp) = 550 ft lbwt/sec = 33,000 ft lbwt/min = 745.7 watts 

1 kilowatt (kw) = 1.341 hp = 737.6 ft lbwt/sec = 0.9483 Btu/sec 

1 nt/m 2 = 10 dynes/cm 2 = 9.869 X lO" 6 atmosphere = 2.089 X 10~ 2 lbwt/ft 2 

1 lbwt/in 2 = 6895 nt/m 2 = 5.171 cm mercury = 27.68 in. water 

1 atmosphere (atm) = 1.013 X 10 5 nt/m 2 = 1.013 X 10 6 dynes/cm 2 = 14.70 lbwt/in 2 
= 76 cm mercury = 406.8 in. water 



Angle 



1 radian (rad) = 57.296°; 1° = 0.017453 rad 



Appendix B 



Astronomical Data 



Mass 



Radius 



Mean density- 



Mean surface gravitational 
acceleration 



Escape velocity 
at surface 



Period of rotation 
about axis 



Universal gravitational 
constant G 



THE SUN 



4.4 X 10 30 lb or 2.0 X 10 30 kg 



4.32 X 10 5 mi or 6.96 X 10 5 km 



89.2 lb/ft 3 or 1.42 gm/cm 3 



896 ft/sec 2 or 273 m/sec 2 



385 mi/sec or 620 km/sec 



25.38 days or 2.187 X 10 6 sec 



1.068 X 10~ 9 
ft 3 /lb-sec 2 



or 6.673 X 10 ~ 8 cm 3 /gm-sec 2 



Mean distance from earth 



Period of rotation 
about earth 



Equatorial radius 



Mass 



Mean density 



Mean surface gravitational 
acceleration 



Escape velocity 



Period of rotation 
about axis 

Orbital speed 



Orbital eccentricity 



THE MOON 



239 X 10 3 mi or 3.84 X 10 5 km 



27.3 days or 2.36 X 10 6 sec 



1080 mi or 1738 km 



1.63 X 10 23 lb or 7.38 X 10 22 kg 



27.3 days 



.055 



208 lb/ft 3 or 3.34 gm/cm 3 



5.30 ft/sec 2 or 1.62 m/sec 2 



1.48 mi/sec or 2.38 km/sec 



or 2.36 X 10 6 sec 



.64 mi/sec or 1.02 km/sec 



342 



APPENDIX Bl 



ASTRONOMICAL DATA 



343 



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Appendix C 



Solutions of Special 
Differential Equations 

DIFFERENTIAL EQUATIONS 

An equation which has derivatives- or differentials of an unknown function is called a 
differential equation. The order of the differential equation is the order of the highest 
derivative or differential which is present. A solution of a differential equation is any 
relationship between the variables which reduces the differential equation to an identity. 

Example 1. 

The equation j*-=2y is a differential equation of first order, or order 1. A solution of this equation 
is y = ce 2x where c is any constant, since on substituting this into the given differential equation we have 

the identity 

2ce 2x = 2ce 2x 

Example 2. 

The equation x 2 dx + y 3 dy = is a differential equation of first order. A solution is x*/3 + y*l± = c 
where c is any constant, since taking the differential of the solution we have 

d(x s /S + y*/4) = or x 2 dx + y z dy - 

Example 3. 

The equation ^| - 3^ + 2y = 4x is a differential equation of second order. A solution is 

CLOG 0/QCr 

y = de x + c 2 e 2x + 2x + 3 since 

&£- S *!L + 2 y = ( Cl e x + 4c 2 e 2x ) - 3( Cl e x + 2c 2 e 2x + 2) + 2(c t e x + c 2 e 2x + 2x + 3) - 4* 
dx 2 dx 

In the above examples we have used x as independent variable and y as dependent 
variable. However, it is clear that any other symbols could just as well have been used. 
Thus, for instance, the differential equation of Example 3 could be 

H - 3^ + 2x = U 
dt 2 dt 

with independent variable t and dependent variable * and solution x = cie* + c 2 e 2t + 2t + S. 

The above equations are often called ordinary differential equations to distinguish them 

a2V r) 2 V 

from partial differential equations such as -^ = c 2 -^" involving two or more independent 
variables. 

ARBITRARY CONSTANTS. GENERAL AND PARTICULAR SOLUTIONS 

In the above examples the constants c, d, c 2 can take on any values and are called 
arbitrary constants. In practice an nth order differential equation will have a solution 
involving exactly n independent arbitrary constants. Such a solution is called the general 
solution. All special cases of the general solution obtained by giving the constants special 
values are then called particular solutions. For instance in Example 3 above if we let 
ci = 6, c 2 = -3 in the general solution y = w* + c 2 e 2 * + 2x + 3, we obtain the particular 
solution y = 5e x - 3e 2x + 2x + 3. 

344 



APPENDIX C] SOLUTIONS OF SPECIAL DIFFERENTIAL EQUATIONS 345 

Particular solutions are often found from certain conditions imposed on the problem 
and sometimes called boundary or initial conditions. In Example 3 for instance, if we wish 
to satisfy the conditions y = 5 when x = and if = dy/dx = 1 when x = 0, we obtain 
ex = 5, c 2 = —3. 

A problem in which we are required to solve a differential equation subject to given 
conditions is often called a boundary-value problem. 

SOLUTIONS TO SOME SPECIAL FIRST ORDER EQUATIONS 

The following list shows some important methods for finding general solutions of first 
order differential equations. 

1. Separation of Variables 

If a first order equation can be written as 

F(x)dx + G(y)dy = (1) 

then the variables are said to be separable and the general solution obtained by direct 
integration is 

J F(x) dx + C G(y) dy =■ c (2) 

2. Linear Equations 

A first order equation is called linear if it has the form 

g + P(x)y = Q(x) (S) 

Multiplying both sides by e J , this can be written 

-^{yJ Pdx } = Qe Spdx 
Then integrating, the general solution is 

y e ) pdx — 1 Qe^ Pdx dx + c 

or y = e~f Pdx C Qef Pdx dx + ceS Pdx (4) 

C p dx 

The factor e J is often called an integrating factor. 

3. Exact Equation 

The equation 

Mdx + Ndy = (5) 

where M and N are functions of x and y is called an exact differential equation if 
Mdx + Ndy can be expressed as an exact differential dU of a function U(x,y). In such 
case the solution is given by U(x, y) = c. 

A necessary and sufficient condition that (5) be exact is 

§M = dN 
By dx K ' 

In some cases an equation is not exact but can be made exact by first multiplying 
through by a suitably chosen function called an integrating factor as in the case of the 
linear equation. 



346 SOLUTIONS OF SPECIAL DIFFERENTIAL EQUATIONS [APPENDIX C 

4. Homogeneous Equation 

If an equation has the form 

P- = f(V-) (7) 

dx \x/ 

it is called a homogeneous equation and can be solved by the transformation y = vx. 

Using this, (7) becomes 

dv r,/ v dx dv , s , 

v + x Tx = F{v) or ~x~ = fW 1 ^ {) 

in which the variables have been separated. Then the general solution is 

fdx = C dv c where v = /x {9) 

J x J F(v) - v 

Occasionally other transformations, which may or may not be evident from the form 
of a given differential equation, serve to obtain the general solution. 



SOLUTIONS OF HIGHER ORDER EQUATIONS 

The following list shows certain equations of order higher than one which can often 
be solved. 

1. -^ = F(x). In this case the equation can be integrated n times to obtain 

dx n v ' 

y = f • • • f F(x) dx n + ci + c 2 x + czx 2 + ••• + e n x n ~ l 

2 SlL = f ( x !^L) . in this case y is missing and if we make the substitution dy/dx = v 
dx 2 \ dx J 

we find that the equation becomes 

dv „. x 

Tx = F(x ' v) 
a first order equation. If this can be solved we replace v by dy/dx, obtaining another 
first order equation which then needs to be solved. 

q ^y = p( v ^}L\ . Here x is missing and if we make the substitution dy/dx = v, 
' dx 2 \ a 'dxJ 

noting also that TO , , , 

d 2 y _ cLv_ _ dv&n _ dv 

dx 2 ~ dx ~ dydx dy' 

the given equation can be written as a first order equation 

which then needs to be solved. 

LINEAR DIFFERENTIAL EQUATIONS OF ORDER HIGHER THAN ONE 

We shall consider solutions of linear second order differential equations. The results 
can easily be extended to linear higher order equations. 
A linear second order equation has the form 

^ + P (x) f x + Q( X )y = IHx) (10) 



APPENDIX C] SOLUTIONS OF SPECIAL DIFFERENTIAL EQUATIONS 347 

If y c is the general solution of the equation 

[obtained by replacing the right hand side of (10) by zero] and if y p is any particular solution 
of (10), then the general solution of (10) is 

y = y c + y p (12) 

The equation (11) is often called the complementary equation and its general solution is 
called the complementary solution. 



LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS 

The complementary solution is easily obtained when P(x) and Q(x) are constants A and 
B respectively. In such case equation (11) can be written 

+ 4f + *" = » W 

If we assume as solution y = e ax where a is constant in (11), we find that a must satisfy 
the equation 

a 2 + Aa + B = 

This equation has two roots, and the following cases arise. 

1. Roots are real and distinct, say a x ¥- a 2 . 

In this case solutions are e aiX and e 012 *. It also follows that c x e a & and c 2 e a * x are solu- 
tions and that the sum c x e a & + c 2 e a * x is the general solution. 

2. Roots are real and equal, say « 1 = « 2 . 

In this case we find that solutions are e aiX and xe aiX and the general solution is 

c x e a ^ x + c 2 xe aiX . 

3. Roots are complex. 

If A and B are real, these complex roots are conjugate, i.e. a + bi and a — bi. In such 
case solutions are e ia+hi)x = e ax e bix = e ax (cos bx + i sin bx) and e (a ~ bi)x (cos bx — i sin bx). 
The general solution can be written e ax (c x cos bx + c 2 sin bx). 

PARTICULAR SOLUTIONS 

To find the general solution of 

we must find a particular solution of this equation and add it to the general solution of (13) 
already obtained above. Two important methods serve to accomplish this. 

1. Method of Undetermined Coefficients. 

This method can only be used for special functions R(x) such as polynomials and the 
exponential or trigonometric functions having the form e px , cos px, sin px where p 
is a constant, together with sums and products of such functions. See Problems C.17 
and C.18. 

2. Method of Variation of Parameters. 

In this method we first write the complementary solution in terms of the constants 
c x and c 2 . We then replace c x and c 2 by functions f t (x) and f 2 (x) so chosen as to satisfy 
the given equation. The method is illustrated in Problem C.19. 



348 SOLUTIONS OF SPECIAL DIFFERENTIAL EQUATIONS [APPENDIX C 

Solved Problems 

DIFFERENTIAL EQUATIONS. ARBITRARY CONSTANTS. 
GENERAL AND PARTICULAR SOLUTIONS 

Cl. (a) Prove that y = ce~ x + x-1 is the general solution of the differential equation 

% ~ x+v = ° 

(b) Find the particular solution such that y = 3 when x - 0. 

(a) If y - ce~ x + x - 1, then dy/dx = —ee~ x + 1 and so 

dy/dx - x + y - (-ce' x + 1) - x + (ce~ x + x - 1) = 

Thus 2/ = ce- x + a; — 1 is a solution; and since it has a number of arbitrary constants 
(namely one) equal to the order of the differential equation, it is the general solution. 

(6) Since y = 3 when x = 0, we have from y = ce~ x + x-1, 3 = c - 1 or c = 4. Thus 
y = 4e~ x + x — 1 is the required particular solution. 

C.2. (a) Prove that x = c^ + c 2 e~ st + sin * is the general solution of 

^ + 2^ - 3* = 2cos* - 4 sin* 
dt 2 at 

(b) Find the particular solution such that x-1, dx/dt = -3 at t = 0. 

(a) From a = Cje* + c 2 e~ 3t + sin t we have 

^ = Cl e* - 3e 2 e- 8t + cost, "^Jf = c i et + 9c 2 e- 3t - sin t 

Then ^ + 2 — - 3a; = (c ie t + 9c 2 e-3t _ sin t ) + 2( Cl e* - 3c 2 e"« + cos t) 

dt2 dt - 3(cje* + c 2 e-3t + sin t) 

= 2 cos t — 4 sin t 

Thus a; = c^* + c 2 e~ 3t + sin t is a solution; and since it has two arbitrary constants while the 
differential equation is of order two, it is the general solution. 

(6) From part (a), letting t = in the expressions for x and dx/dt, we have 

2 = c t + c 2 fci + c 2 = 2 

or i . 

-3 = Cl - 3c 2 + 1 L c i _ 3c 2 = _4 

Solving, we find c x — 1/2, c 2 = 3/2. Then the required particular solution is 

x = ^e* + f e~ 3t + sin t 

SEPARATION OF VARIABLES 

C.3. (a) Find the general solution of (x + xy 2 ) dx + (y + £ 2 2/) dy = 0. 

(b) Find the particular solution such that y = 2 when a; = 1. 

(a) Write the equation as x(l + y 2 ) dx + y(l + x 2 ) dy = 0. Dividing by (1 + x 2 )(l + y 2 ) ¥> to 
separate the variables, we find 

xdx ydy _ Q ^) 

1 + a;2 ^ 1 + y2 



Then we have on integrating, 



J xdx , f ydy 
1 + x 2 J 1 + y 

1 In (1 + a; 2 ) + £ In (1 + ?/ 2 ) = *i 



^y* = Cl 



APPENDIX C] SOLUTIONS OF SPECIAL DIFFERENTIAL EQUATIONS 349 

This can be written ^ In {(1 + x 2 )(l + y 2 )} - c x or 

(1 + X 2 ){1 + yl) = C2 (j g) 

which is the required general solution. 

(6) Since y = 2 when x = 1, we have on substitution in (2), c 2 = 10; thus the required particular 
solution is 

(1 + x 2 )(l + y 2 ) = 10 or x 2 + y 2 + x 2 y 2 = 9 

C.4. Solve ^ = RH 2 if R = 1 when t = 1. 
at 

dR 
Separating the variables, we have -^ - t 2 dt. Integrating both sides, 

1 t 3 

~R = 3 +C 

Substituting t = 1, R = 1 we find c = -4/3. Thus 

1 t 3 4 „ 3 

— or it 



R 3 3 4 - t3 



LINEAR EQUATION 

dv 
C.5. Solve -5— + 2xy = x 3 + x if 2/ = 2 when x = 0. 

This is a linear equation of the form (S), page 345, with P = 2x, Q = X s + x. An integrating 

I 2.x dix 2 

factor is e J — e x . Multiplying the given equation by this factor, we find 

e x2 ^- + 2xye x2 = (x 3 + x)e x2 
ax 

j 

which can be written t~{v e x2 ) — (# 3 + x)e x2 

ax 

Integrating, y e x = I (x 3 + x)e x2 dx + c 

or, making the substitution v = x 2 in the integral, 

y e* 2 = -1x20*2 + c 

Thus y = ±x 2 + ce~ x2 

Since y = 2 when x = 0, we find c = 2. Thus 

y = \x 2 + 2e~ x2 

Check: If y = \x 2 + 2e~ x2 , then dy/dx = x - 4xe~ x2 . Thus 

^-+ 2xy = x - 4xe~ x2 + 2x(\x 2 + 2e~* 2 ) = x 3 + x 



C.6. Solve ~=SU + lifU = when t = 0. 
dt 



Writing the equation in the form 

dU 



dt~* U = 1 W 



350 SOLUTIONS OF SPECIAL DIFFERENTIAL EQUATIONS [APPENDIX C 

we see that it is linear with integrating factor e-* ~ 3dt = e~ st . Multiplying (1) by e" 8 ', it can be 

written as 

4i(Ue-M) = e -*t 
at 

Integrating, we have Ue~ 3t = — ^e~ 3t + c 

Since U = when t = 0, we obtain c = £. Thus 

Ue-zt = _£ e -3t + |. or u = £(e 3t -l) (2) 

Another method. The equation can also be solved by the method of separation of variables. Thus 
we have , r . 

dU = dt 



3U + 1 

Integrating, £ In (3t7 + 1) = t + c 

Since U = when t = 0, we find c = so that £ In (BU + 1) = t. Thus *7 = £(e 3t -l). 



EXACT EQUATIONS 

C.7. Solve (Sx 2 + y cos x) dx + (sin x - 4y 3 ) dy = 0. 

Comparing with M dx + N dy = 0, we have M = Sx 2 + y cos x, N = sin g - 4s/ 3 . Then 
3Af/3i/ = cos x = dN/dx and so the equation is exact. Two methods of solution are available. 

Method 1. 

Since the equation is exact, the left side must be an exact differential of a function U. By 
grouping terms, we find that the equation can be written 

3a 2 dx + (y cos x dx + sin x dy) — Ay 3 dy = 
i.e., d(x s ) + d(y sin *) + d(-y±) = or d(» 3 + y sin x - y 4 ) = 

Integration then gives the required solution, X s + y sin x — y 4 = c. 

Method 2. 

The given equation can be written as 

(3a; 2 + y cos x) dx + (sin x - 4j/ 3 ) dy = dU = -^dx + —dy 

Then we must have 

ATT STJ 

(1) ^ - 3a;2 + 2/ cos a; (2) — = sin x - 4j/ 3 

v ' dx oy 

Integrating (1) with respect to x, keeping y constant, we have 

U = X s + y sin x + F(y) 

Then substituting this into (2), we find 

sin a; + F'(y) = sin x - 4?/ 3 or F'(y) = -4^ 

where F'(y) = dF/dy. Integrating, omitting the constant of integration, we have F(y) = -y* so 

that 

U = X s + y sin x — y* 

Then the given differential equation can be written 

dU = d(x* + y sin x — y 4 ) = 
and so the solution is x z + y sin x — y 4 = c. 



APPENDIX C] SOLUTIONS OF SPECIAL DIFFERENTIAL EQUATIONS 351 

HOMOGENEOUS EQUATIONS 

C.8. Solve p- = e» /x + %-. 
dx x 

Let y = vx. Then the equation can be written 

dv , dv 

v + %-7- = e v + v or %~j~ — e 

dx Ax 

Separating the variables, — = e~ v dv. Integrating, Ins = — e~ v + c. Thus the general solution is 

lnsc + e~ y,x — c. 

SOLUTIONS OF HIGHER ORDER EQUATIONS 

C.9. Solve ^ = 1 + cos t where U = 2, dU/dt = 3 at t = 0. 

Integrating once, 

dU/dt = t + sin* + c x 

Then since dU/dt = 3 at * = 0, we find e t - 3. Thus 

dU/dt = t + sin* + 3 

Integrating again, U = ^t 2 — cos t + St + c 2 

Now since U = 2 at t - 0, we find c 2 = 3. The required solution is 

U = \& - cos t + 3t + 3 

CIO. Solve a#" + 2y' = x 2 where y' = dy/dx, y" = d?y/dx 2 . 

Since y is missi'ig, let y' = dy/da; = v. Then the equation can be written 

(i) *g+2* = * 2 or (f) £+§* = x 

in i> with integrating factor J (2/x) dx = e 2 ln * = c ln * 2 = * 2 . Multiplying 

A 
dx 

Then by integration, a5 2 v = x*/4 + c x or 

v = dy/dx = x*/4 + cjx* 

Integrating again, y = a^/12 — c x lx + c 2 . 

Cll. Solve yy" + {y'f = where y' = dy/dx, y" - d 2 y/dx 2 . 

Since x is missing, let y' ~ dy/dx = v. Then 

„ _ dv __ dv^djt _ v dv^ 
V ~ dx dy dx dy 

and the given equation can be written 

yv ^ + v2 = ° or v ( v % + v ) = ° 

/ x dv , _ 

so that (1) v = or (2) y-^ + v = 

From (1), y'=0 or 2/ = ^. From (2), — + -^ = 0, i.e. lnv + lny = c 2 or ln (vy) = c 2 so 
that vy = c 3 and 



This is a linear equation 
(2) by 05 2 , it can be written as 

(x 2 v) = x 



352 SOLUTIONS OF SPECIAL DIFFERENTIAL EQUATIONS [APPENDIX C 

v = dy/dx — c 3 /y or y dy = c 3 dx 

Integrating, y 2 /2 = c 3 x + c 4 or y 2 = Ax + B 

Thus solutions are y = c t and y 2 — Ax + B. Since the first is a special case of the second, the 
required general solution can be written y 2 = Ax + B. 

LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS 

C.12. Solve 4^-4^-52/ = 0. 
dx 2 dx 

Letting y = e ax in the equation, we obtain 

(a 2 - 4a — h)e ax = or a 2 — 4a - 5 = 

Thus (a — 5)(a + 1) = and a = 5,-1. Then solutions are e 5x and e~ x and the general solution is 

y = c x e* x + c 2 e~ x . 

C.13. Solve 0+10^ + 252/ = 0. 

Letting y = e ax , we find a 2 + 10a + 25 = 0, i.e. (a + 5)(a + 5) = 0, or a = -5, -5. Since 
the root is repeated, solutions are e~ 5x and xe~ 5x . Then the general solution is y = c x e~^ x + c 2 xe~ 5x . 

C.14. Solve ^§ + 4^ + 4x = 0. 
at 2 dt 

Letting x = e at , we find a 2 + 4a + 4 = or a = -2, -2. Then the general solution is 
x = c^" 2 * + c 2 te~ 2t = «~ 2t («i + c 2 t). 

C.15. Solve + 2 g + 5v = 0. 

Letting 2/ = e ax , we find a 2 + 2a + 5 = or a = -1 ± 2t. Then solutions are e (_1 + 2i)x - 
e -x e 2ix = e -x( cos 2a; + t sin2a;) and e (-1 - 2i)a; = e -a: e- 2te = e-z(cos2ar — i sin2a;). The general 
solution is y = e~ x (c x cos 2x + c 2 sin2x). 

C.16. Solve d 2 y/dx 2 + <* 2 y - 0. 

Letting y = e ax , we find a 2 + w 2 = or a = ± ia. Then solutions are e to * = cos wo; + i sin w« 
and e -i6U! = cos <*x — i sin ax. The general solution is thus y = c t cos ax + c 2 sin v>x. 

METHOD OF UNDETERMINED COEFFICIENTS 
C.17. Solve 0-4||-5i/ = x 2 + 2e* 

By Problem C.12 the complementary solution, i.e. the general solution, of 

&-«£-* = ° 

is y c = c t e& + c 2 e~ x (1) 

Since the right side of the given equation contains a polynomial of the second degree (i.e. x 2 ) 
and an exponential (2e 3x ), we are led to the trial particular solution 

y p = Ax 2 + Bx + C + De* x (*) 

where A, B, C, D are constants to be determined. 

Substituting (2) for y in the given equation and simplifying, we find 

(2A - 4B - BO + (-8A - 5B)x - SAs 2 - 8De 3 * = a 2 + 2e^ x 

Since this must be an identity, we must have 



APPENDIX C] SOLUTIONS OF SPECIAL DIFFERENTIAL EQUATIONS 353 

2A-4B-5C = 0, -8A - 55 = 0, -5A = 1, -8D = 2 
Solving, we find A = -£, B=£,C = -{%, Z> = -j. Then from (2), 

2/p — 5 X T 25 X 125 4 e 

Thus the required general solution of the given equation is 
which can be checked by direct substitution. 

C.18. Solve 0+lO^| + 252/ = 20 cos 2x. 

The complementary solution [by Problem C.13] is 

y c - Cl e- 5x + c 2 xe- 5x CO 

Since the right side has the term cos 2x, we are led to the trial solution 

y p = A cos 2x + B sin 2x (2) 

Substitution into the given equation yields, after simplifying, 

(21A + 205) cos 2x + (215 - 20A) sin 2x = 20 cos 2x 
Equating coefficients of like terms, we have 21A + 205 = 20, 215 - 20/1 = 0. Solving, we find 
A = 84i> B = 84i so that the P articular solution is 

420 n i 400 . o 

y v = I_t cos 2* + -57T sin 2a 



'P 841 



and the general solution of the given equation is 



V = Vc + V P = c i e ~ Sx + We~s* + fff cos 2x + 



420 ___ o„ , 400 

841 



sin 2x 



METHOD OF VARIATION OF PARAMETERS 
C.19. Solve d 2 y/dx 2 + y = tana:. 

The complementary solution is as in Problem C.16 with u = 1: 

y c = c_ cos x + c 2 sin a CO 

We now assume that the solution to the given equation has the form 

y = fi cos x + / 2 sin x (2) 

where /_ and / 2 are suitable functions of a;. From (*) we have, using primes to denote differentiation 
with respect to x, 

dy/dx = -fi sin * + f 2 cos x + /_ cos * + / 2 sin (5) 

Before finding d 2 y/dx 2 let us observe that since there are two functions / t and / 2 to be determined 
and only one relation to be satisfied [namely that the given differential equation must be satisfied] 
we are free to impose one relation between f x and / 2 . We choose the relation 

/i cos x + / 2 sin x = (4) 

so as to simplify (8) which then becomes 

dy/dx = — /1 sin x + f 2 cos a (5) 

Another differentiation then leads to 

d 2 y/dx 2 = -fi cos * - / 2 sin x - f[ sin x + / 2 cos as (0) 

From (2) and (0) we see that the given differential equation can be written 

dty/dx* + y - -fi sin x + f 2 cos x = tan x (7) 

Thus — /{ sin x + / 2 cos x = tan x (5) 

From (.4) and (5) we find fi = - sin 2 x/cos x, / 2 = sin x. Thus 



354 SOLUTIONS OF SPECIAL DIFFERENTIAL EQUATIONS [APPENDIX C 

C sin 2 x , f 1 - cos 2 x , C . , , 

f. = — I ax = — I ax = — I (sec * — cos x) ax 

J1 J cos* J cos x J 

= — In (sec x + tan a;) + sin a? + c x 

f 2 = \ sin x dx = — cos * + c 2 

Substituting in (2) we find the required general solution, 

y = d cos x + c 2 sin a; — cos a; In (sec x + tan x) 



Supplementary Problems 

DIFFERENTIAL EQUATIONS. ARBITRARY CONSTANTS. 
GENERAL AND PARTICULAR SOLUTIONS 

C.20. Check whether each differential equation has the indicated general solution. 

(a) ^f - 2^ + x = t ; x = (c t + c 2 t)e* + t + 2 
at'- at 

(*>) *4r + V = &; t?-StU = c 
at 

C.21. (a) Show that z = e _t (Ci sin t + c 2 cost) is a general solution of 

(6) Determine the particular solution such that z = — 2 and dz/dt = 5 at t = 0. 
Ans. (6) « = e -t (3 sin * — 2 cos t) 

SOLUTIONS TO SPECIAL FIRST ORDER EQUATIONS 

C.22. Solve dy/dx = -2xy if y = 4 when x = 0. An*. # = 4e- a:2 



C.23. Solve 



dz = ty/T^z* A ^^—^_^ r —- 



dt 



Vi - * 2 



C.24. Solve x^--2y = x if y(l) = 5. Arts, y = 6a; 2 - x 

C.25. Solve (a; + 2y) dx + (2a; - hy) dy = if 1/ = 1 when a; = 2. Ans. a; 2 + 4»y - 5y 2 - 7 

C.26. Solve ^ = - + ^a- Ans. \nx + (x/y) = c 

dx x x z 

C.27. Solve (ye x — e~v) dx + (xe~y + e x ) dy = 0. Ans. ye x — xe~ y - c 

C.28. Solve (x + xy) dx + (xy + y)dy = 0. Ans. (x + l)(y + 1) = ce* + y 

C.29. Show that the differential equation {Ay - a; 2 ) dx + xdy = has an integrating factor which de- 
pends on only one variable and thus solve the equation. Ans. xty — ^x 6 = c 

SOLUTIONS OF HIGHER ORDER EQUATIONS 

C.30. Solve d?U/dt z = t + e~* if U = 3, dU/dt = 2 at t = 0. Ans. U = ±t? + e~* + 3t + 2 

C.31. Solve x J^ - 3 ^ = a; 2 . Ans. y = -^x» + c x x* + c 2 



APPENDIX C] SOLUTIONS OF SPECIAL DIFFERENTIAL EQUATIONS 355 

C.32. Solve ^ff+ 2 (^) 2 = °' Ans. US = Cl t + c 2 



C.33. Solve 



H*)"]' -(£)'• ^-<*-^ +<*-*>* = ! 



LINEAR HIGHER ORDER DIFFERENTIAL EQUATIONS 

C.34. Solve |^-2^-82/ = 0. Arts, y = c x e* x + c^- 2 * 

C.35. Solve ^? + 4^+ W = if U = l,dU/dt = at t = 0. Ans. U = (l + 2t)e- 2 * 

dt £ at 

C.36. Solve ^| + 4^J+ 5z = 0. Ans. z = e~ 2t (ci cos t + c 2 sin*) 

ac z at 

C.37. Solve 4^| + 25y = if 2/(0) = 10, y'(0) = 25. Ans. y = 10(cos§05 + sulfa) 

C.38. Solve0-4|-2/ = 0. Ans. y = e 2x ( Cl e^ x + c 2 e~^ x ) 

C.39. Solve 4y" - 20y' + 25y = 0. Ans. (c t + c 2 x)e 5x/2 

C.40. Solve + 3^ + 2y = «-«*. Ans. y = Cl e~* + c 2 e~^ + \e~* x 

C.41. Solve ^ + ^r~ 2U = 6t - 10 cos 2* + 5. 
at 2 at 

Ans. U = c x e- 2t + c 2 e* - St - 4 + | sin 2t - § cos 2t 

C.42. Solve y" + y — sec « by use of the method of variation of parameters. 
Ans. y = c x sin x + c 2 cos x + x sin a; — cos a; In sec x 

C.43. Solve (a) Problem C.40 and (6) Problem C.41 by variation of parameters. 

C.44. Solve each of the following equations by any method. 

(a) y" - 5y' + 6y = 50 sin 4a; (c) y" + 4y = esc 2* 

(6) y" + 2y' - 3y = xe~ x (d) 1/" + 82/' + 25y = 25x + 33 + 18e"* 

A«s. (a) y = c x e 2x + c 2 e Sx + 2 cos 4* — sin 4a? 

(6) ^ = Cl e x + c 2 e~ Sx — \xe~ x 

(c) y = c t cos 2x + c 2 sin 2a; - %x cos 2x — \ sin 2a; In (esc 2a:) 

(^ y = e -4x( Cl cos 3* + c 2 sin 3a;) + x + 1 + e _x 

C.45. Solve y" + 4j/' + 4y = e" 2 *. Aw*, y = (e t + c 2 a;)e~ 2a; + \xH~^ 

C.46. Solve simultaneously: dx/dt + y = e*, x — dy/dt = t. 
Ans. x = c t cos t — c 2 sin t + ^e* + t, 
y = d sin t + c 2 cos £ + £e* — 1 

C.47. Solve 2/" + 2/ = 4 cos t if 2/(0) = 2, y'(0) ~ -1. Is the method of undetermined coefficients ap- 
plicable? Explain. Ans. y = 2 cos t - sin t + 2t sin £ 

C.48. Show how to solve linear equations of order higher than two by finding the general solutions of 

(a) y"' - 6y" + 1W ~ 62/ = 36a;, (6) 2/ (iv) + W + y = x*. 

Ans. (a) 2/ = W + C2 e2x + c 3 e3x - 6* - 11 

(6) y = d cos * + c 2 sin x + x{c 3 cos x + c 4 sin a;) + a; 2 — 4 



Appendix D 



Index of Special Symbols 
and Notations 



The following list shows special symbols and notations used in this book together with 
the number of the page on which they first appear. All bold faced letters denote vectors. 
Cases where a symbol has more than one meaning will be clear from the context. 



Symbols 

a length of semi-major axis of ellipse or hyperbola, 38, 118, 119 

a n Fourier cosine coefficients, 196 

a acceleration, 7 

a P ip acceleration of particle P 2 relative to particle P v 7 

A area, 122 

A vector potential of electromagnetic field, 309 

A areal velocity, 122 

cA amplitude of steady state oscillation, 90 

cA max maximum amplitude of steady state oscillation, 90 

6 length of semi-minor axis of ellipse or hyperbola, 38, 118, 119 

b n Fourier sine coefficients, 196 

B magnetic intensity, 83 

B unit binormal, 7 

c speed of light, 54 

C curve, 6 

D F ,D M time derivative operator in fixed and moving systems, 144 

e lt e 2 , e 3 unit vectors, 72 

E total energy, 36 

E electric intensity, 84 

f force due to friction, 65 

f vX internal force on particle v due to particle X, 173 

/ frequency, 89 

f a frequencies, 316 

F force, 33 

F 12 force of particle 1 on particle 2, 33 

F av average force, 60 

F D damping force, 87 

F v impulsive forces, 285 

F„ (a) , F„ (c) actual and constraint forces acting on particle v, 170 

f a generalized impulse, 285 

f v force (external and internal) acting on particle v of a system of particles, 168 

356 



APPENDIX D] INDEX OF SPECIAL SYMBOLS AND NOTATIONS 357 

g acceleration due to gravity, 62 

G gravitational constant, 120 

Q generating function, 314 

h Planck's constant, 338 

H Hamiltonian, 311 

Jl Hamiltonian under a canonical transformation, 314 

i unit vector in direction of positive x axis, 3 

/ moment of inertia, 225 

I c moment of inertia about axis through center of mass, 226 

I xx , I yy , I zz moments of inertia about x, y, z axes, 254 

I xy , I yz , I xz products of inertia, 254 

I U I 2 ,I 3 principal moments of inertia, 255 

J) v impulses, 285 

$ angular impulse, 228 

j unit vector in direction of positive y axis, 3 

J a phase integral or action variable, 316 

k unit vector in direction of positive z axis, 3 

K radius of gyration, 225 

I length, 90 

L Lagrangian, 284 

£ Lagrangian under a canonical transformation, 314 

m mass, 33 

m rest mass, 54 

M total mass of a system of particles, 166 

n number of degrees of freedom, 282 

n orbital quantum number, 338 

N normal component of reaction force, 65 

N number of particles in a system, 166 

N unit normal, 7 

p a generalized or conjugate momenta, 284 

p momentum, 33 

P period, 89 

P a new generalized momenta under a canonical transformation, 314 

<? power, 34 

q electrical charge, 84 

q a generalized coordinates, 282 

Q a new generalized coordinates under a canonical transformation, 314 

r spherical coordinate, 32 

r position vector or radius vector, 4 

f position vector of center of mass, 166 

r x unit vector in radial direction, 25 

r{, position vector of particle v relative to center of mass, 169 

R radius of curvature, 8 

R range, 75 

R max maximum range, 75 

R resisting force, 64 

R resultant of forces, 47 

% rigid body, 228 

% Routh's function or Routhian, 337 

s arc length, 7 
s spin angular velocity, 269 
S generating function, 316 
gj" generating function depending on old position coordinates and new momenta, 314 



358 INDEX OF SPECIAL SYMBOLS AND NOTATIONS [APPENDIX D 

t time, 6 

T kinetic energy, 35 

T tension, 74 

T unit tangent vector, 7 

T generating function depending on old and new position coordinates, 323 

V generating function depending on new coordinates and old momenta, 334 
v lim limiting speed, 70 

v raax) v min maximum and minimum orbital speeds, 143 

v velocity, 7 

v velocity of center of mass, 167 

Vp 2 /P i velocity of particle P 2 relative to particle P lt 7 

\' v velocity of particle v relative to center of mass, 169 

v i2> v i2 relative velocities of particles along common normal before and after impact, 194 

V potential or potential energy, 35 

"U generating function depending on old and new momenta, 334 

w a angle variables, 316 

W work, 34 

W weight, 62 

y c complementary solution, 347 

y p particular solution, 347 

V transverse displacement of vibrating string, 195 
z cylindrical coordinate, 32 

Z atomic number, 338 



Greek Symbols 



a angle made by vector with positive x direction, 24 

a index of summation, 282 

a angular acceleration, 29 

/? angle made by vector with positive y direction, 24 

/? damping constant, 88 

f3 ratio of speed of particle to speed of light, i.e. vie, 54 

7 angle made by vector with positive z direction, 24 

8 logarithmic decrement, 89 
8 variation symbol, 313 

S a p Kronecker delta, 336 

A determinant, 336 

e coefficient of restitution, 195 

£ eccentricity, 118 

e cylindrical coordinate, 32 

e Euler angle, 257 

e polar coordinate, 25 

6 spherical coordinate, 32 

9 t unit vector perpendicular to radial direction, 25 

k curvature, 8 

k spring constant, 86 

X colatitude, 152 

^v ^2> ^3> • • • Lagrange multipliers, 280, 284 

A v A 2 , A 3 components of torque along principal axes, 256 

A torque or moment, 36 

A c torque or moment about center of mass, 229 



APPENDIX D] INDEX OF SPECIAL SYMBOLS AND NOTATIONS 359 

H coefficient of friction, 65 

fi reduced mass, 182 

v index of summation, 166 

p cylindrical coordinate, 32 

p density in phase space, 318 

a density, 114 

a torsion, 31 

t radius of torsion, 31 

t time, 81 

t volume, 166 

<f> Euler angle, 257 

<p phase angle or epoch, 88 

spherical coordinate, 32 

* scalar potential, 309 

4> a generalized force, 283 

a angular speed, 8 

« lt <o 2 , « 3 components of angular velocity along principal axes, 256 

u angular velocity, 144 

a x , fi,,, a z components of angular momentum along x, y, z axes, 254 

a lt « 2 » n 3 components of angular momentum along principal axes, 255 

Q angular momentum, 37 

Notations 

|A| magnitude of A, 4 

AB magnitude of distance from A to B, 11 

A • B dot or scalar product of A and B, 4 

A X B cross or vector product of A and B, 5 

A • (B X C) scalar triple product, 5 

A X (B X C) vector triple product, 5 

A(w) vector function of u, 6 

A(x,y,z) vector function of x,y,z, 8 

0(w) scalar function of u, 6 

<p(x, y, z) scalar function of x, y, z, 8 

A, A time derivatives of A, i.e. dA/dt, d 2 A/dt 2 , 8 

A(tt) du indefinite integral of A(m), 6 

A(m) du definite integral of A(w), 6 

C 

I integral along curve C, 9 

<s> integral around a closed path, 9 

V del operator, 8 

V<£ = grad <f> gradient of <p, 8 

V • A = div A divergence of A, 8 

V X A = curl A curl of A, 9 

f(r) magnitude of central force, 116 

[F, G] Poisson bracket of F and G, 331 



/ 



INDEX 



Absolute motion, 34, 60 
Acceleration, 1, 7, 17-20 

along a space curve, 8, 20 

angular, 8, 145, 148 

apparent, 149 

centrifugal, 145 

centripetal, 8, 20, 21, 150 

Coriolis, 145, 150 

due to gravity, 62 

in cylindrical coordinates, 32 

in moving coordinate systems, 
145, 146, 148, 149 

in polar coordinates, 26 

in spherical coordinates, 32 

instantaneous, 7 

linear, 145 

normal, 7, 8, 19, 20 

relative, 7, 18, 19 

tangential, 7, 8, 19, 20 

true, 149 

uniform, 62, 65, 66 
Action and reaction, 33 
Action integral, 313 
Action variables, 316, 328 
Actual force, 170 
Air resistance, 63, 69-72 
Amplitude, 86, 87 

modulation, 102 

of damped oscillatory motion, 88 

of steady-state oscillation, 90 
Angle variables, 316, 328, 329 
Angular acceleration, 8, 145, 148 
Angular impulse, 170, 228, 237 
Angular momentum, 37, 45-47 

about principal axes, 255 

conservation of, 37, 45-47, 168, 228, 237 

of a rigid body, 227, 236, 254, 259, 260 

of a system of particles, 168, 169, 176, 179 

of the earth about its axis, 150 

principle of, 227, 229, 236, 238 

relationship to torque, 37, 46, 
168, 169, 176 
Angular speed, 8 (see also Angular velocity) 
Angular velocity, 144, 148 

in terms of Euler angles, 258 

of rigid body, 253 
Anharmonic oscillator, 115 
Aphelion, 120 
Apogee, 120 

Approximations, method of successive, 154, 159 
Apsides, 143 
Arbitrary constants, 344, 348 



Arc length, 7 

Areal velocity, 122, 123 

Area, of parallelogram, 15 

of triangle, 31 
Areas, law of, 116, 123 
Associative law, 3, 10 

for rotations, 245 
Astronomical data, 342, 343 
Astronomy, definitions in, 119, 120 
Asymptotes, 119 

Attraction, 120, 121, 129-133, 136 
Atwood's machine, 76, 305 
Axioms, 1 

Binomial theorem, 106 

Binormal, 7, 8 

Bob, of simple pendulum, 90 

Body axes, 257 

Body centrode or locus, 229, 240, 241 

Body cone, 257, 266 

Bohr's quantum theory, 338 

Boundary value problems, 195, 345 

Bound vectors, 9, 10 

Brachistochrone problem, 322 

Calculus of variations, 313, 320-333 

connection with Hamilton's principle, 321 
Canonical coordinates, 314 
Canonical equations, Hamilton's, 311 
Canonical transformations, 314, 323-325 

condition for, 314 
Celestial mechanics, 311 
Center, of ellipse, 118 

of force, 116 

of gravity, 167 

of hyperbola, 119 
Center of mass, 166, 172-175, 183-186 

motion of, 167 

motion of system relative to, 169, 178, 179 
Central field, 116 (see also Central force) 

equations of motion for particle in, 

116, 122, 123 

potential energy of particle in, 

117, 123-125 

Central force, 116, 121, 122, 168, 318 
(see also Central field) 

determination of, from orbit, 118, 125-127 
Centrifugal acceleration, 145 

force, 146 
Centripetal acceleration, 8, 20, 21, 150 

force, 146, 150 
Centrode, space and body, 229, 240, 241 



361 



362 



INDEX 



Centroid {see Center of mass) 
Cgs system, 33, 62, 339, 340 
Chain, hanging, 186, 187 

sliding, 83 
Characteristic determinant, 198 

frequencies, 198 
Chasle's theorem, 224, 275 
Circular motion, 8, 20, 21, 95, 96 
Clock, 2 

Coefficient of friction, 65 
Colatitude, 152 
Collinear vectors, 23 
Collisions of particles, 194, 195, 200-202 
Comet, 121 

Commutative law, for dot and cross 
products, 3, 5, 10, 13, 14 
for Poisson brackets, 331 
for rotations, 230, 231 
Complementary equation and solution, 347 
Components, of a vector, 4 
Compound pendulum, 228, 237, 238, 279, 291 
Compression time, 194 
Conical pendulum, 157 
Conic section, 118, 127, 128 
Conjugate momentum, 284, 288 
Conservation, of angular momentum, 
37, 45-47, 168, 228, 237 
of energy, 36, 43-45, 124, 
169, 227, 229, 236, 240 
of momentum, 37, 167, 173 
Conservative force fields, 35, 43-45, 283, 
286, 287 
condition for, 35, 50, 51 
Constant of the motion, 312 
Constrained motion, 64, 65, 72, 73 
Constraint force, 170 
Constraints, 170, 180 
holonomic and non-holonomic, 170, 180, 

283, 284, 286, 287 
reaction due to, 64, 65 
Contact transformations (see Canonical 

transformations) 
Continuous functions, piecewise, 197 
Continuous systems of particles, 165, 195 
Conversion factors, table of, 341 
Coordinate systems, 3, 4 
inertial, 34, 39 
moving, 144-164 
non-inertial, 144 
Coplanar vectors, condition for, 16 
Coriolis acceleration, 145, 150 

force, 146 
Cosines, direction, 24 

law of, 27 
Couples, 226, 227, 235 
Critically damped motion, 88, 96, 98, 99 
Cross products, 5, 13-15 

determinant expression for, 14, 15 
distributive law, 5, 14 
failure of commutative law for, 5, 13, 14 
Curl, 8, 9, 21, 22 
of gradient, 9, 29 



Curvature, 8, 20 

radius of, 8, 20 
Cycle, 87 

Cyclic coordinates, 312 
Cycloid, 84, 85, 106, 107, 302, 322, 323 
Cycloidal pendulum, 112, 303 
Cyclones, 163 

Cylinder, vibrations of, 104, 105 
Cylindrical coordinates, 32 

acceleration in, 32 

gradient in, 61 

Lagrange's equations in, 291, 292 

velocity and acceleration in, 32 

D'Alembert's principle, 171, 182, 229 

Damped harmonic oscillator, 87, 88, 96-99 

Damped oscillatory motion, 88, 98 

Damping coefficient, 88 

Damping forces, 64, 87 

Deceleration, 29 

Decrement, logarithmic, 89, 97, 98 

Definite integrals, 6 

Definitions, 1 

Deformable bodies, 165 

Degrees of freedom, 165, 172, 224, 225, 253, 282 

of a rigid body, 172, 253, 259 
Del, 8 
Density, 165 

in phase space, 312 
Derivatives, in moving coordinate systems, 
144, 145, 147, 148 
notation for time, 8 
of vectors, 6, 16, 17 
Determinant, characteristic, 198 
Jacobian, 337 
secular, 198, 215 
Dextral system, 4 
Diagonal, main or principal, 254 
Difference equation, 216 
Difference of vectors, 3 

Differential and derivative operators, 8, 144, 148 
Differential equations, 344-355 

partial, 195, 344 
Differential, exact, 51, 52 
Dimensions, 2, 339, 340 
Direct collision, 194 
Direction, 2 

cosines, 24 
Directrix, 118 

Dirichlet conditions, 197, 206, 207 
Discontinuities, 197, 204, 207 
Discrete system of particles, 165 
Displacement, 2, 224 
true, 170 
virtual, 170 
Dissipative forces, 64 
Distance, 2 

between two points, 11 
Distributive law, 3 

for Poisson brackets, 332 
Divergence, 8, 21, 22 

of curl, 9, 29 
Dot products, 4, 5, 12, 13 



INDEX 



363 



Dot products (cont.) 

commutative law for, 5 

distributive law for, 5, 12 
Double pendulum, 285, 286, 299-301 
Drumhead, vibrating, 195 
Dumbbell, 278 
Dynamics, 1 
Dyne, 33 

Earth, flat, 63 
motion of particle relative to, 145 
rotation of, 150, 257, 265 
Eccentricity, 118 

Einstein's laws of relativity, 34, 61 
Elastic bodies, 165 
Elastic collisions, perfectly, 195, 201 
Elastic constant, 86 
Elasticity, 194 

modulus of, 86 
Elastic string, vibrations of 

(see Vibrating string) 
Electrical charge, 83, 84 
Electromagnetic field, 84, 309 
Hamiltonian for particle in, 338 
Lagrangian for particle in, 309 
Ellipse, 38, 104, 118, 119, 121, 127, 128 
Ellipsoid of inertia, 255, 256, 263, 264 
Elliptic functions, 106, 272, 279 

integrals, 106, 108 
Energy, conservation of, 36, 43-45, 124, 169, 
227, 229, 236, 240 
kinetic, (see Kinetic energy) 
of simple harmonic oscillator, 87, 99 
potential (see Potential energy) 
total, 36 
English systems, 63, 339 
Epoch, 87, 88, 93 
Equality of vectors, 2 
Equilibrant, 47 
Equilibrium, 37, 171 

in a uniform gravitational field, 65, 66, 74, 75 
of a particle, 37, 38, 47, 48 
of a rigid body, 229, 241, 242 
of a system of particles, 170, 180, 181 
position, 86 

stable, 38, 48, 49, 60, 141, 171, 230 
Escape speed or velocity, 134 
Euclidean geometry, 1, 2 
Euler angles, 253, 257, 267, 268, 301 
angular velocity in terms of, 258 
Euler's equations of motion, 256, 264 

from Lagrange's equations, 302 
Euler's or Lagrange's equations, 313, 320 
Euler's theorem, 224 

on homogeneous functions, 305, 306, 317 
Even extension of a function, 208 
Even functions, 196 
Event, 2 
E volute, 112 
Exact differential, 51, 52 
Exact differential equation, 345, 350 
Extremal, 313 
Extremum or extreme value, 313 



Field, scalar or vector, 8 
Flat earth, 63 (see also Earth) 
Focus, 118 
Force, 33 

axiomatic definition of, 33, 49 
centrifugal and centripetal, 145, 146, 150 
constraint, 170 
damping, 64, 87 
generalized, 283 
units of, 33, 339, 340 
Forced vibrations, 89, 99-102 
resonance and, 90, 100, 101 
Force fields, conservative (see Conservative 
force fields) 
non-conservative, 37, 47 
uniform, 62, 65, 66 
Foucault pendulum, 146, 154-156 
Fourier coefficients, 196, 206 
Fourier series, 195-197, 203-208 
convergence of, 197 
half range, 197, 207, 208 
Fourier coefficients for, 196, 206 
solution of vibrating string by 
(see Vibrating string) 
Fps system, 33, 62, 339, 340 
Frames of reference, Newtonian, 33, 34 
Freely falling bodies, 63, 67 
Free vectors, 9, 10 
Frenet-Serret formulas, 31 
Frequencies, characteristic, 198 
Frequency, fundamental, 211 
natural, 89, 98 

obtained by Hamiltonian methods, 316, 329 
of precession, 257, 265, 270, 273, 274 
of resonance, 90 

of simple harmonic motion, 86, 87 
Friction, 65 
coefficient of, 65 
motion involving, 73 
Fss system, 63, 339, 340 
Function, scalar and vector, 8 
Fundamental frequency, 211 

Generalized coordinates, 282, 285, 286 

forces, 283 

impulse, 285 

momenta, 284, 288 

velocities, 283 
General solution of differential equation, 

344, 348 
Generating functions, 314, 315, 323-325 
Geometry, Euclidean, 1, 2 
Gimbal, of a gyroscope, 258 
Gradient, 8, 21, 22 

curl of, 9, 29 

in cylindrical and spherical coordinates, 61 
Gram, 33 
Gravitation, universal law of (see Universal 

law of gravitation) 
Gravitational constant, universal, 120 
Gravitational potential, 120, 121, 133, 143 
Gravitational system of units, 63, 339 



364 



INDEX 



Gravity, 62 

center of, 167 

vibrating string under, 214, 215 
Gyration, radius of, 225 
Gyrocompass, 278 
Gyroscopes, 258, 268-273 

Half range Fourier sine and cosine series 

197, 207, 208 
Hamiltonian, 311, 317, 318, 330 

for conservative systems, 311 

for particle in electromagnetic field, 338 
Hamilton-Jacobi equation, 315, 325-327, 330, 331 

for Kepler's problem, 326, 327 

for one dimensional harmonic oscillator, 
325, 326 

solution of, 315, 316 
Hamilton's equations, 311, 317, 318 
Hamilton's principle, 313, 320-323 
Harmonic oscillator, damped, 87, 88, 96-99 

simple, 86-90, 92-102 

two and three dimensional, 91, 103, 104 
Herpolhode, 257, 266 
Holonomic, 170, 180, 283, 284, 286, 287 
Homogeneous equation, 346, 351 
Homogeneous functions, 305 

Euler's theorem on, 305, 306, 317 
Hooke's law, 86 

Hyperbola, 104, 119, 121, 127, 128 
Hyperbolic functions, 54 

Ignorable coordinates, 312, 315 

Impact, 194 

Impulse, 36, 45-47, 169, 170, 180 

angular, 170, 228, 237 

generalized, 285 

relation to momentum, 36 
Impulsive forces, 285, 295-298 
Inclined plane, 64, 65, 72 

motion of particle down, 72, 73 

motion of sphere down, 239, 240 

projectile motion on, 75, 76, 81 
Incompressible fluid, 313 
Indefinite integrals of vectors, 6 
Independence of path, 9 

condition for, 50, 51 
Inelastic collisions, perfectly, 195, 201 
Inertial frames of reference, 33, 34, 39 

classical principle of relativity for, 39 
Inertial system, 34, 39 
Initial point, of a vector, 2 
Instability, 38 
Instantaneous, acceleration, 7 

axis of rotation, 224, 225, 229 

center of rotation, 225, 229, 240, 241 

power, 34 (see also Power) 

velocity, 7 
Integral equations, 154 
Integrals of vectors, 6, 16, 17 

line (see Line integrals) 
Integrating factor, 345, 351 
Internal forces, 177, 178 
Invariable line and plane, 256, 257, 266 



Invariant, 34 

Isolation, of a system, 64 

Iteration, method of, 154, 159 

Jacobian determinant, 337 

Kepler's laws, 120, 128, 129, 223 

deduction of from Newton's universal 
law of gravitation, 125, 126, 129 
Kilogram, 33 

weight, 63 
Kinematics, 1 
Kinetic energy, 34, 35, 41-43 

about principal axes, 255 

in terms of Euler angles, 258, 268 

in terms of generalized velocities, 283, 287, 288 

of a rigid body, 227, 236, 259, 260 

of a system of particles, 168, 169, 179, 182 

of rotation, 229 

of translation, 229 

relationship to work, 35, 41, 169 

relativistic, 55 
Kronecker delta, 336 

Lagrange multipliers, 280, 284, 292, 295 
Lagrange's equations, 282-310, 320 

and calculus of variations (see Calculus 
of variations) 

for conservative systems, 284, 288-292 

for non-conservative systems, 284 

for non-holonomic systems, 284, 285, 
292-295, 303 

with impulsive forces, 285, 295-298 
Lagrangian function, 284, 311 

for particle in electromagnetic field, 309 
Latitude, 152 
Laws, 1 

Lemniscate, 138 
Length, 2 

Light, speed of, 34, 54 
Limiting speed or velocity, 70, 72 
Line, 1 

of action of a vector, 10 
Linear equations, 345-347, 349, 350, 352 
Linear harmonic oscillator, 86 (see also 

Harmonic oscillator) 
Linear impulse (see Impulse) 
Linear momentum (see Momentum) 
Line integrals, 9, 22, 23 

evaluation of, 22, 23 

independence of path of, 9, 23 
Liouville's theorem, 312 

proof of, 318-320 
Lissajous curves or figures, 91 
Logarithmic decrement, 89, 97, 98 
Lorentz force, 84 

Magnetic field, 83 

Main diagonal, of moment of inertia 

matrix, 254 
Major axis, of ellipse, 118 
of hyperbola, 139 



INDEX 



Mass, 2, 33 

axiomatic definition of, 49 
center of (see Center of mass) 
changing, 194 

of the earth, 129 
reduced, 182, 231 
rest, 54, 61 
units of, 33 
Mathematical models, 1 
Matrix, moment of inertia, 254 
Matter, 1, 2 
Mechanics, 1 

relativistic, 34 
Membrane, vibrating, 195 
Meteorite, 121 
Metric system, 339 
Minor axis, of ellipse, 119 

of hyperbola, 139 
Mks system, 33, 62, 339, 340 
Mode of vibration, normal, 194 
Models, mathematical, 1 
Modulation, amplitude, 102 
Modulus of elasticity, 86 
Moment, of couple, 226 

of force, 36 

of momentum, 37 (see also Angular momentum) 
Momental ellipsoid, 256 
Moments of inertia, 225, 231-233, 254, 259, 
260, 263, 264 

matrix, 254 

principal (see Principal moments of inertia) 

special, 226 

theorems on, 225, 233-235 
Momentum, 33, 167 

angular (see Angular momentum) 

conjugate, 284, 288 

conservation of, 37, 167, 173 

generalized, 284, 288 

moment of, 37 (see also Angular momentum) 

of a system of particles, 167, 169, 172, 173 

principle of, 238 
Momentum coordinates, 312 
Moon, 119, 342 

Multiply-periodic vibrations, 194 
Musical note, 211 

Natural frequency and period, 89, 98 
Newton, 33 

Newtonian frames of reference, 34 
Newton's collision rule, 194, 202 

laws of motion, 33-41 

universal law of gravitation (see 
Universal law of gravitation) 
Nodes, line of, 257 

Non-holonomic, 170, 180, 283, 284, 286, 287 
Non-inertial systems, 144, 145 
Normal frequencies, 194, 198 

for a double pendulum, 300, 301, 308 

for a vibrating string, 210, 211 

for a vibrating system of particles, 215, 216 
Normal modes of vibration, 194, 198, 199 

for a vibrating string, 210, 211 
Normal, principal, 7, 8, 20 



Normal (cont.) 

to a surface, 24 
Null vector, 3 
Nutation, 270, 272 

Oblique collisions, 194 

Odd extension, of a function, 207 

Odd functions, 196 

Operators, derivative, 144 

Optics, 335 

Orbit, 116, 120 

determination of from central force, 
117, 118, 125-127 

weightlessness in, 135, 136 
Order of a differential equation, 344 
Oscillations, forced, 89 (see also 

Forced vibrations) 
Oscillator, anharmonic, 115 

harmonic (see Harmonic oscillator) 
Overdamped motion, 88, 98, 99 
Overtones, 211 

Pappus, theorems of, 193 
Parabola, 63, 104, 119, 121, 127, 128 

as curve of motion of projectile, 68 
Paraboloid of revolution, 107, 108 
Parachutist, motion of, 69, 70 
Parallel axis theorem, 226, 233, 234 
Parallelepiped, volume of, 5, 15, 16 
Parallelogram, area of, 5, 15 
Parallelogram law, 2 
Partial differential equation, 195, 344 

of vibrating string (see Vibrating string) 
Particles, 2 

equilibrium of, 37, 38 

systems of, 165-193 

vibrations of, 194, 197-199 
Particular solutions, 344, 345, 347, 348 
Path, independence of, 9, 50, 51 
Pendulum, bob, 90 

compound, 228, 237, 238, 279, 291 

conical, 157 

cycloidal, 112, 303 

double, 285, 286, 299-301 

Foucault, 146, 154-156 

seconds, 110 

simple (see Simple pendulum) 
Perigee, 120 
Perihelion, 120 
Period, 53 

natural, 89 

of damped motion, 89 

of harmonic oscillator, 87 

of motion in a magnetic field, 83 

of simple harmonic motion, 86, 87 

of simple pendulum, 91, 105, 106 

orbital, 135, 136 

sidereal, 120 
Perpendicular axes theorem, 226, 234, 235 
Phase, angle, 87, 88, 93 

integrals, 316, 328, 329 

out of, 93 

space, 312, 318-320 



366 



INDEX 



Piano string, vibrations of, 195 
(see also Vibrating string) 
Piecewise continuous functions, 197 
Planck's constant, 338 
Planets, 119, 343 
Poinsot's construction, 257 
Point, 1, 2 

Poisson bracket, 331, 332 
Polar coordinates, 25, 26 

gradient in, 54 

velocity and acceleration in, 26 
Polhode, 257, 266 
Position, 2 

coordinates, 312 

vector, 4 
Potential, 35 (see also Potential energy) 

relation to stability, 38 

scalar, 35, 309 

vector, 309 
Potential energy, 35, 36, 43-45 
(see also Potential) 

in a central force field, 117, 123-125 

in a uniform force field, 64, 69 

of a system of particles, 169, 176-178 

principle of minimum, 230 

relation of to work, 35, 44 
Pound, 33 

weight, 63 
Poundal, 33 
Power, 34, 41-43, 227, 237 

relation to work, 42 
Precession, 156, 256, 270, 272 

frequency of, 257, 265, 270, 273, 274 
Principal axes of inertia, 255, 260-263 
Principal diagonal, 254 
Principal moments of inertia, 255, 260-263 

method of Lagrange multipliers for, 280 
Principal normal, unit, 7, 8, 20 
Products of inertia, 254, 259, 260 
Products of vectors, by a scalar, 3 

cross (see Cross products) 

dot (see Dot products) 
Projectiles, 62, 63 

maximum height of, 68 

motion of, 68, 69, 71, 72 

on an inclined plane, 75, 76, 81 

range of (see Range of projectile) 

time of flight, 68 
Pulley, 76, 289, 290 

Quantum mechanics, 311 
Quantum number, orbital, 338 
Quantum theory, 338 

Radius, of curvature, 8, 20 

of gyration, 225 

of torsion, 31 
Radius vector, 4 
Range of projectile, 68 

maximum, 69 

on inclined plane, 75, 76 

on rotating earth, 164 
Reaction, 33 



Reaction (cont.) 

due to constraints, 64, 65 
Rectangular coordinate systems, 3, 4 

right handed, 4 
Reduced mass, 182, 231 
Reference level, 64 
Relative acceleration, 7, 18, 19 

velocity, 7 
Relativistic mechanics, 34 
Relativity, classical principle of, 34, 39 

Einstein's laws of, 34, 61 

theory of 54, 55, 61, 143, 337 
Representative point, 312 
Resistance, air, 63, 69-72 
Resisting forces, 64 

Resisting medium, motion in, 64, 69-72 
Resonance, 90, 100, 101 
Restitution, coefficient of, 195 
Restitution time, 194 
Rest mass, 54, 61 
Restoring force, 86 
Resultant of vectors, 2 
Rheonomic, 283, 286, 287 
Right handed system, 4 
Rigid bodies, 165, 170, 224, 230, 231, 236 

equilibrium of, 229, 241, 242 

force free motion of, 256, 257, 265 

general motion of, 224, 253, 259 

motion of, about a fixed axis, 236 

plane motion of, 224-252 

symmetric, 257 
Rockets, 173, 194, 199, 200 

motion of, 199, 200 
Rotating coordinate systems, 144, 147, 148 
Rotation, 224, 253 

associative and commutative laws for, 230, 
231, 245 

finite, 230, 231 

of the earth, 150, 257, 265 

pure, 253 
Routh's function or Routhian, 337 

Satellites, 119 
Scalar function, 8 
Scalar potential, 35 

for electromagnetic field, 309 
Scalar product (see Dot product) 
Scalar triple product, 5 
Scalars, 2 

Scleronomic, 283, 286, 287 
Seconds pendulum, 110 
Secular determinant, 198, 215 
Semi-major and minor axes, 118, 119, 129 
Separation of variables, 210, 316, 345, 347, 348 
Sidereal period, 120 
Simple closed curve, 9 
Simple harmonic motion, 86 

(see also Simple harmonic oscillator) 

amplitude, period and frequency of, 86, 87 
Simple harmonic oscillator, 86-90, 92-102 

amplitude, period and frequency of, 86, 87 

damped, 87, 88 

energy of, 87, 99 



INDEX 



367 



Simple harmonic oscillator (cont.) 
forced vibrations of, 89, 99-102 
resonance and, 90, 100, 101 
Lagrange's equations for, 306 
Simple pendulum, 86, 87, 90, 91, 102, 103 

length of equivalent, 228 
Sines, law of, 27 
Sliding vector, 9, 10 
Slug, 63 

Solar system, 119 

Solution of differential equation, 344 
Space, 1, 2 
Space axes, 257 

Space centrode or locus, 229, 240, 241 
Space cone, 257, 266 
Special relativity (see Relativity) 
Speed, 7 (see also Velocity) 
angular, 8 
escape, 134 
of light, 34 
orbital, 135, 136 
Sphere, particle sliding down, 76, 77, 82 

sphere rolling down, 244, 303, 304 
Spherical coordinates, 32 
gradient in, 61 
Lagrange's equations in, 306 
velocity and acceleration in, 32 
Spin, 270, 272 
Spring constant, 86 
Spring, vibrations of, 86, 93-95 
Stability of equilibrium, 38, 48, 49, 60, 141, 

171, 230 
Stable point, 38 
Star, 119 
Statics, 1, 37, 38, 47 

in a uniform gravitational field, 65, 66, 74, 75 
of a particle, 37, 38, 47, 48 
of a rigid body, 229, 241, 242 
of a system of particles, 170, 180, 181 
Statistical mechanics, 311 
Steady-state solution, 89 
Stiffness factor, 86 
Sum of vectors, 2 

obtained graphically and analytically, 12, 48 
Sun, 119, 342 

Superposition principle, 199 
Surface, normal to, 24 
Symmetric matrix or tensor, 254 
Systems of particles, 165-193 

Tangent vector, unit, 7, 8, 19 

Tautochrone problem, 113 

Tension, 74, 76 

Tensor, moment of inertia, 254 

Terminal point, of a vector, 2 

Theorems, 1 

Time. 1, 2 

principle of least, 335 
Top, 258, 268-273, 274 

Lagrange's equations for motion of, 301, 302 

motion of, 258, 268-273 

sleeping, 274 



Top (cont.) 

steady precession of, 270 
Torque, 36, 45-47, 168, 176 

of a couple, 226 

relation to angular momentum, 37, 46, 168, 
169, 176 
Torsion, 31 

constant of, 308 

radius of, 31 
Transformation equations, 282, 285, 286 

canonical, 314, 323-325 
Transient solution, 89 
Translation, 224, 253 
Transverse vibration of a string 

(see Vibrating string) 
Triple products, 5, 15, 16 

scalar, 5 

vector, 5 
Two and three body problems, 121, 223 

Underdamped motion, 88, 98 
Undetermined coefficients, method of, 347, 

352, 353 
Uniform acceleration, 62, 65, 66 

force field, 62, 65, 66 
Uniformly accelerated motion, 62, 65, 66 
Units, 2, 339-341 
Unit vectors, 3 

rectangular, 3, 4 
Universal law of gravitation, 120, 128, 129 

deduction from Kepler's laws, 128, 129 
Unstable equilibrium, 171, 230 

Variation of an integral, 321 

Variation of parameters, method of, 347, 353, 354 

Variation symbol, 313, 337 

Variations, calculus of 

(see Calculus of variations) 
Vector algebra, laws of, 3, 10-12 
Vector field, 8 
Vector function, 8 
Vector potential, 309 
Vector product (see Cross product) 
Vector triple product, 5 
Vectors, 1, 2 

algebra of, 2, 3 

bound, 9, 10 

components of, 4 

definition of, 2 

free, 9, 10 

magnitude of, 11, 13 

sliding, 9, 10 
Velocity, 1, 6, 7, 17-19 

angular (see Angular velocity) 

apparent, 148 

areal, 122, 123 

escape, 134 

generalized, 283 

in cylindrical coordinates, 32 

in moving cordinate systems, 145, 148, 149 

in polar coordinates, 26 

in spherical coordinates, 32 



368 



INDEX 



Velocity (cont.) 

instantaneous, 7 

limiting, 70, 72 

of a rigid body, 253, 259 

relative, 7 

true, 148 
Vertices, of ellipse, 118 

of hyperbola, 119 
Vibrating string, 195, 202, 203, 209-212 

considered as a system of particles, 215-217 

under gravity, 214, 215 
Vibrating systems of particles, 194, 197-199 
Vibrations, of a cylinder, 104, 105 

forced (see Forced vibrations) 
Violin string, vibrations of, 195 

(see also Vibrating string) 
Virtual displacements, 170 
Virtual work, Lagrange's equations and, 292 



Virtual work, principle of, 170, 229 

Weight, 62 
apparent, 162 

Weightlessness, 135, 136 

Work, 34, 41-43, 168, 169, 176-178, 237 
generalized forces and, 283, 287, 288 
in rotation of a rigid body, 227 
relationship of to kinetic energy, 168, 169, 

176, 177 
relationship of to potential energy, 44 
virtual (see Virtual work) 

x direction, 4 

y direction, 4 

z direction, 4 
Zero vector, 2